Let $f(x,y)=x^3+x(y^3+1) +e^{y-1}$. Let $(x_0,y_0)=(2,1)$. 4 parts to the question
(a) Compute $\frac{\partial{f}}{\partial{x}}(2,1)$ and $\frac{\partial{f}}{\partial{y}}(2,1)$ \begin{aligned} \end{aligned}
(b)compute $f(2,1)$\begin{aligned} \end{aligned}
(c) Use a linear approxomation to estimate the change in $f$ if $x$ is increased by $0.2$ and $y$ is increased by $0.3$\begin{aligned} \end{aligned}
(d) Suppose that $x$ is increased by $0.2$. Use a linear approximation to estimate how much y should be decreased so that $f$ remains the same.\begin{aligned} \end{aligned}
Part (a):
\begin{aligned}\frac{\partial{f}}{\partial{x}}=3x^2 + y^3 + 1 \end{aligned}
\begin{aligned}\frac{\partial{f}}{\partial{y}}=3xy^2-e^{y-1}\end{aligned}
\begin{aligned}\frac{\partial{f}}{\partial{x}}(2,1)=3*(3)^2 + 1^3+1 = 29\end{aligned}
\begin{aligned}\frac{\partial{f}}{\partial{y}}(2,1)=2(1)^2-e^{1-1}=1\end{aligned}
Part (b):
\begin{aligned}f(2,1)= 2^3+2(1^3+1)+e^{1-1}=13\end{aligned}
Part (c):\begin{aligned} L(x,y)= f(2,1) + f_x(2,1)(x-2) +f_y(2,1)(y-1)\end{aligned}
\begin{aligned} L(x,y)= 13 + 29(x-2) + 1(y-1) \end{aligned}
\begin{aligned} L(x,y)= 13 + 29x - 58 + y - 1 \end{aligned}
\begin{aligned} L(x,y)= 29x + y - 46 \end{aligned}
\begin{aligned} L(2.2,1.3) = 29(2.2)+ 1.3 -46 = 19.1\end{aligned}
part (d) \begin{aligned} 19.1=29(2.2) + 1.3 - a - 46 \end{aligned}
just solve for a and you have the answer.
Am I correct in these solutions?
 A: In part (a) you wrote:
$$\frac{\partial f}{\partial x} = 3x^3 + y^3 + 1$$
$$\frac{\partial f}{\partial y}= 3xy^2 - e^{y - 1}$$
But it should be:
$$\frac{\partial f}{\partial x} = 3x^2 + y^3 + 1$$
$$\frac{\partial f}{\partial y}= 3xy^2 + e^{y - 1}$$

(b) is correct and in (c) and (d) I think they want you to use the formula:
$$\Delta f\approx \frac{\partial f}{\partial  x} \Delta x + \frac{\partial f}{\partial y} \Delta y \ \ \ (1)$$
Try to use this and then edit you question and show what you did, and then I can see if you made it correctly.

In part (C) you got the partial derivatives at $(2,1)$ wrong because you used the wrong formula for the darivatives:
$$f_x(2,1) = 3 \cdot 2^2 + 1^3 + 1 = 14$$
$$f_y(2,1) = 3\cdot 2 \cdot 1^2 + e^{1-1}=7$$
I don't quite understand what you tried to do with that $L$ function and why you defined it the way you did. In the exercise they said: "$x$  is increased by $0.2$ and $y$ is increased by $0.3$". This means that $\Delta x = 0.2$ and $\Delta y = 0.3$. Now, If you just plug these values in formula (1) you get:
$$\begin{align}
\Delta f & \approx f_x(2,1) \Delta x +  f_y(2,1) \Delta y\\
\\
& \approx 14 \cdot 0.2 + 7 \cdot 0.3 = 4.9
\end{align}$$
So you conclude that: $\Delta f \approx 4.9$ and this is what they asked you to calculate.

In (D) You just do the same but you solve for $\Delta y$:
You want $\Delta y = 0$ because you don't want it do change, so you solve:
$$\begin{align}
f_x(2,1) \cdot 0.2 +  f_y(2,1) \Delta y = 0
\end{align}$$
and so you end up with $\Delta y = -0.4$
And this is the answer that you want.
