Answer
\[
\begin{array}{c}
1=f_{x}(0,-1)
\end{array}
\]
Work Step by Step
Given that $f(x, y)$ is differentiable at (0,-1) and partial derivatives exist, we use the Local Linear Approximation Theorem:
\[
f\left(x_{0}, y_{0}\right)+f_{x}\left(x_{0}, y_{0}\right)\left(x-x_{0}\right)+f_{y}\left(x_{0}, y_{0}\right)\left(y-y_{0}\right)=L(x, y)
\]
So:
$ \Delta x= -x_{0}+x$, $\Delta y=-y_{0}+y$, $x_{0}=0, y_{0}=-1,x=0.1, and \ y=-1.1 $
Thus:
\[
\begin{array}{c}
f_{y}(0,-1)=-2 \quad \& \quad f(0,-1)=3 \\
L(x, y)=L(0.1,-1.1)=3.3 \\
L(x, y)=f\left(x_{0}, y_{0}\right)+f_{x}\left(x_{0}, y_{0}\right)\left(x-x_{0}\right)+f_{y}\left(x_{0}, y_{0}\right)\left(y-y_{0}\right) \\
\Rightarrow 3.3=3+(-2)(-0.1)+f_{x}(0,-1)(0.1-0) \\
\frac{0.1}{0.1} =f_{x}(0,-1) \\
1=f_{x}(0,-1)
\end{array}
\]
