Answer
$\frac{∂f}{∂x}=\frac{-2y}{(xy+1)^{3}}$
$\frac{∂f}{∂y}=\frac{-2x}{(xy+1)^{3}}$
$\frac{∂f}{∂x}|_{(1,-1)}$ and $\frac{∂f}{∂y}|_{(1,-1)}$ are not defined at $(1,-1)$
Work Step by Step
We find the partial derivatives as follows:
$f(x,y)=\frac{1}{(xy+1)^2}=(xy+1)^{-2}$
$\frac{∂f}{∂x}=-2(xy+1)^{-3}(y)=-2y(xy+1)^{-3}=\frac{-2y}{(xy+1)^{3}}$
$\frac{∂f}{∂y}=-2(xy+1)^{-3}(x)=-2x(xy+1)^{-3}=\frac{-2x}{(xy+1)^{3}}$
$\frac{∂f}{∂x}|_{(1,-1)}$ and $\frac{∂f}{∂y}|_{(1,-1)}$ are not defined because $f(x,y)$ is not defined at $(1,-1)$
$f(1,-1)=\frac{1}{(1(-1)+1)^2}=\frac{1}{0}$
