Answer
\[
\begin{aligned}
0.5=f_{y}(1,1) &
\end{aligned}
\]
Work Step by Step
Given that $f(x, y)$ is differentiable at (1,1) and the partial derivatives exist, we use the Local Linear Approximation theorem:
\[
L(x, y)=f\left(x_{0}, y_{0}\right)+f_{x}\left(x_{0}, y_{0}\right)\left(x-x_{0}\right)+f_{y}\left(x_{0}, y_{0}\right)\left(y-y_{0}\right)
\]
So:
$\Delta x=-x_{0}+x$, $\Delta y=-y_{0}+y$, $x_{0}=1, y_{0}=1$ and $x=1.1, y=0.9$
Thus:
\[
\begin{aligned}
2=f_{x}(1,1) & \\
3=f(1,1) & \\
L(x, y) &=L(1.1,0.9)=3.15 \\
\Rightarrow 3.15 &=3+2(1.1-1)+f_{y}(1,1)(0.9-1) \\
0.5=f_{y}(1,1) &
\end{aligned}
\]
