The question is really asking for a tangent plane, so lets first find partial derivatives and then plug in the point.

$\displaystyle f(x,y)=6+\frac{x^2}{4}+\frac{y^2}{16}$, $\displaystyle f(-2,4)=6+\frac{(-2)^2}{4}+\frac{(4)^2}{16}=6+1+1=8$

$\displaystyle f_x(x,y)=\frac{x}{2}$, $\displaystyle f_x(-2,4)=\frac{-2}{2}=-1$

$\displaystyle f_y(x,y)=\frac{y}{8}$, $\displaystyle f_y(-2,4)=\frac{4}{8}=\frac{1}{2}$

Remember that we need to build the linear approximation general equation which is as follows.

$\displaystyle L(x,y)\approx f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$

$\displaystyle L(x,y)\approx 8-(x+2)+\frac{1}{2}(y-4)$

$\displaystyle L(x,y)\approx 8-x-2+\frac{1}{2}y-2$

$\displaystyle L(x,y)\approx 4-x+\frac{1}{2}y$

To find the equation of the tangent plane, we use the formula

$\displaystyle z- z_0 = f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$.

Taking partial derivatives, we have

$\displaystyle f_x(x,y) =$ $\displaystyle 2y-e^{-x}$

$\displaystyle f_y(x,y) = 2x$

Substituting our $\displaystyle x, y$ values into these, we get

$\displaystyle f_x(0,1) =1$

$\displaystyle f_y(0,1) = 0$

Substituting our point into $\displaystyle x_0, y_0, z_0$, and partial derivative values in the formula we get

$\displaystyle z - 1 = 1(x-0)+0(y-1)$

$\displaystyle z - 1 = x$

$\displaystyle -x+z =1$.

We are just asking for the equation of the tangent plane:

Step 1: Find $\displaystyle f(x,y)$

$\displaystyle f(x,y)=2+\frac {(-3)^2}{9}+\frac {2^2}{4}=2+\frac {9}{9}+\frac {4}{4}=2+1+1=4$

Step 2: Take the partial derivative of $\displaystyle \frac {x^2}{9}$ with respect with (x,y):

$\displaystyle \partial(\frac {x^2}{9})=\frac {2x}{9}$

Step 3: Evaluate the partial derivative of x at $\displaystyle -3$

$\displaystyle \frac {2(3)}{9}=\frac {6}{9}=\frac {2}{3}$

Step 4: Take the partial derivative of $\displaystyle \frac {y^2}{4}$ with respect to $\displaystyle (x,y)$:

$\displaystyle \partial (\frac{y^2}{4})=\frac {2y}{4}$

Step 5: Evaluate the partial derivative at $\displaystyle y=2$

$\displaystyle \frac {2(2)}{4}=\frac {4}{4}=1$.

Step 6: Convert (x,y) back into binomials:

$\displaystyle x\rightarrow (x+3)$
$\displaystyle y\rightarrow (y-2)$

Step 7: Write the equation of the tangent line:

$\displaystyle 4+\frac {2}{3}(x+3)+(y-2)$

To find the equation of the tangent plane, we find: $\displaystyle f_x,f_y,z_0$ and evaluate $\displaystyle f_x, f_y$ at the point given. $\displaystyle f_x=\frac{\partial }{\partial x}(2x^2+4y^3)=4x$, $\displaystyle f_y=\frac{\partial }{\partial y}(2x^2+4y^3)=12y^2$, and $\displaystyle z_0=2(1^2)+4(5^3)=502$. Evaluating $\displaystyle f_x, f_y$ at the point $\displaystyle (1,5)$ gets us $\displaystyle f_x=4, f_y=300$. We then plug these values into the formula for the tangent plane: $\displaystyle z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$. We then get $\displaystyle z-502=4(x-1)+300(y-5)$. The equation of the plane then becomes, through algebra, $\displaystyle z=4x+300y-1002$

To find the equation of the tangent plane, we find: $\displaystyle f_x,f_y,z_0$ and evaluate $\displaystyle f_x, f_y$ at the point given. $\displaystyle f_x=\frac{\partial }{\partial x}(3\sin(x)+2(\cos(y))=3\cos(x)$, $\displaystyle f_y=\frac{\partial }{\partial y}(3\sin(x)+2\cos(y))=-2\sin(y)$, and $\displaystyle z_0=3\sin(\frac{\pi}{2})+2\cos(\frac{\pi}{6})=3+\sqrt3$. Evaluating $\displaystyle f_x, f_y$ at the point $\displaystyle (\frac{\pi}{2},\frac{\pi}{6})$ gets us $\displaystyle f_x=0, f_y=-1$. We then plug these values into the formula for the tangent plane: $\displaystyle z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$. We then get $\displaystyle z-(3+\sqrt3)=0(x-\frac{\pi}{2})+-1(y-\frac{\pi}{6})$. The equation of the plane then becomes, through algebra, $\displaystyle z=-y+\frac{\pi}{6}+3+\sqrt3$

To find the equation of the tangent plane, we need 5 things:$\displaystyle f_x,f_y,z(x,y),f_x(x,y),f_y(x,y)$

$\displaystyle f_x=\frac{\partial }{\partial x}(3x^2y)=6xy$

$\displaystyle f_x(2,4)=6*2*4=48$

$\displaystyle f_y=\frac{\partial }{\partial y}(3x^2y)=3x^2$

$\displaystyle f_y(2,4)=3(2)^2=12$

$\displaystyle z(2,4)=3(2)^2(4)=48$

Using the equation of the tangent plane

$\displaystyle z-z(x,y)=f_x(x,y)(x-x_0)+f_y(x,y)(y-y_0)$, we get

$\displaystyle z-48=48(x-2)+12(y-4)$

Through algebraic manipulation to get z by itself, we get

$\displaystyle z=48x+12y-96$

The first thing we need to do is find the partial derivative in respect to $\displaystyle x$, and $\displaystyle y$.

$\displaystyle \frac{\partial}{\partial x}=20x$, $\displaystyle \frac{\partial}{\partial y}=-6y+10$

We need to find the critical points, so we set each of the partials equal to $\displaystyle 0$.

$\displaystyle 0=20x\rightarrow x=0$

$\displaystyle 0=-6y+10 \rightarrow y=\frac{5}{3}$

We only have one critical point at $\displaystyle f\Big(0, \frac{5}{3}\Big)$, now we need to find the function value in order to see if it is inside or outside the disk.

$\displaystyle f\Big(0, \frac{5}{3}\Big)=10(0)^2-3\Big(\frac{5}{3}\Big)^2+10\Big(\frac{5}{3}\Big)=\frac{25}{3}\approx8.33333$

This is within our disk.

We now need to take a look at the boundary, $\displaystyle x^2+y^2=16$. We can solve for $\displaystyle x^2$, and plug it into $\displaystyle f$.

$\displaystyle x^2+y^2=16$

$\displaystyle x^2=16-y^2$

$\displaystyle g(y)=10(16-y^2)-3y^2+10y=-13y^2+10y+160$

We will need to find the absolute extrema of this function on the range $\displaystyle -4\leq y \leq 4$. We need to find the critical points of this function.

$\displaystyle g'(y)=-26y+10$

$\displaystyle 0=-26y+10 \rightarrow y=\frac{5}{13}$

The function value at the critical points and end points are:

$\displaystyle g(-4)=-13(-4)^2+10(-4)+160=-88$

$\displaystyle g(4)=-13(4)^2+10(4)+160=-8$

$\displaystyle g\Big(\frac{5}{13}\Big)=-13\Big(\frac{5}{13}\Big)^2+10\Big(\frac{5}{13}\Big)+160=\frac{2105}{13}\approx161.92$

Now we need to figure out the values of $\displaystyle x$ these correspond to.

$\displaystyle y=-4: x^2=(-4)^2-16 \rightarrow x^2=0\rightarrow x=0$

$\displaystyle y=4: x^2=4^2-16 \rightarrow x^2=0\rightarrow x=0$

$\displaystyle y=\frac{5}{13}: x^2=\Big(\frac{5}{13}\Big)^2-16 \rightarrow x^2=\frac{2679}{169}\rightarrow x\approx\pm\ 3.98$

Now lets summarize our results as follows:

$\displaystyle g(-4)=-88 \rightarrow f(0,-4)=-88$

$\displaystyle g(4)=-8 \rightarrow f(0,4)=-8$

$\displaystyle g\Big(\frac{5}{13}\Big)=\frac{2105}{13} \rightarrow f\Big(-\sqrt{\frac{2679}{169}},\frac{5}{13}\Big)=\frac{2105}{13}$

$\displaystyle g\Big(\frac{5}{13}\Big)=\frac{2105}{13} \rightarrow f\Big(\sqrt{\frac{2679}{169}},\frac{5}{13}\Big)=\frac{2105}{13}$

From this we can conclude that there is an absolute minimum at $\displaystyle (0,-4)$, and two absolute maximums at $\displaystyle \Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ and $\displaystyle \Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$.

First we need to set up our system of equations.

$\displaystyle 2=2\lambda x \rightarrow x=\frac{1}{\lambda}$

$\displaystyle -5=2\lambda y \rightarrow y=-\frac{5}{2 \lambda}$

$\displaystyle x^2+y^2=144$

Now lets plug in these constraints.

$\displaystyle (\frac{1}{\lambda})^2+(-\frac{5}{2\lambda})^2=144$

 $\displaystyle \frac{1}{\lambda ^2}+\frac{25}{4\lambda^2}=144$

$\displaystyle \frac{29}{4\lambda^2}=144$

Now we solve for $\displaystyle \lambda$

$\displaystyle \frac{29}{4\lambda^2}=144\rightarrow \lambda^2=\frac{29}{576}\rightarrow\lambda=\pm\sqrt{\frac{29}{576}}$

If

 $\displaystyle \lambda=\sqrt{\frac{29}{576}}$

$\displaystyle x=\frac{1}{\sqrt{\frac{29}{576}}}\approx 4.46$, $\displaystyle y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx -11.14$

If

 $\displaystyle \lambda=-\sqrt{\frac{29}{576}}$

$\displaystyle x=\frac{1}{\sqrt{\frac{29}{576}}}\approx -4.46$, $\displaystyle y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx 11.14$

Now lets plug in these values of $\displaystyle x$, and $\displaystyle y$ into the original equation.

$\displaystyle f(4.46,-11.14)=2(4.46)-5(-11.14)=64.62$

$\displaystyle f(-4.46,11.14)=2(-4.46)-5(11.14)=-64.62$

We can conclude from this that $\displaystyle f(4.46,-11.14)$ is a maximum, and $\displaystyle f(-4.46,11.14)$ is a minimum.

Let $\displaystyle g = x^2 +2y^2$To find the absolute minimum value, we must solve the system of equations given by

$\displaystyle \triangledown f = \lambda\triangledown g, g =1$.

So this system of equations is

$\displaystyle f_x = \lambda g_x$, $\displaystyle f_y = \lambda g_y$, $\displaystyle g =1$.

Taking partial derivatives and substituting as indicated, this becomes

$\displaystyle 2x = \lambda(2x), 2y = \lambda(4y), x^2+2y^2 = 1$.

From the left equation, we see either $\displaystyle x=0$ or $\displaystyle \lambda = 1$. If $\displaystyle x=0$, then substituting this into the other equations, we can solve for $\displaystyle y, \lambda$, and get $\displaystyle y = \pm \sqrt{2}/2$, $\displaystyle \lambda = 1/2$, giving two extreme candidate points at $\displaystyle (0, \frac{\sqrt{2}}{2}), (0, -\frac{\sqrt{2}}{2})$.

On the other hand, if instead $\displaystyle \lambda =1$, this forces $\displaystyle y = 0$ from the 2nd equation, and $\displaystyle x = \pm 1$ from the 3rd equation. This gives us two more extreme candidate points; $\displaystyle (-1,0),(1,0)$.

Taking all four of our found points, and plugging them back into $\displaystyle f$, we have

$\displaystyle f(-1,0) = 1, f(1,0) = 1, f(0,\frac{\sqrt{2}}{2}) = \frac{1}{2}, f(0,-\frac{\sqrt{2}}{2}) = \frac{1}{2}$.

Hence the absolute minimum value is $\displaystyle \frac{1}{2}$.