Answer
$$\frac{\partial f(2,-2)}{\partial x}=-\frac{1}{4}$$
$$\frac{\partial f(2,-2)}{\partial y}=\frac{1}{4}$$
Work Step by Step
To solve both partial derivatives we will use quotient rule.
The partial derivative with the respect to $x$ is:
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}\left(\frac{xy}{x-y}\right)=\frac{(x-y)\frac{\partial }{\partial x}(xy)-xy\frac{\partial }{\partial x}(x-y)}{(x-y)^2}=\frac{(x-y)\cdot y-xy\cdot 1}{(x-y)^2}=\frac{xy-y^2-xy}{(x-y)^2}=-\frac{y^2}{(x-y)^2}$$
because $y$ is treated as a constant.
The partial derivative with the respect to $y$ is:
$$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}\left(\frac{xy}{x-y}\right)=\frac{(x-y)\frac{\partial}{\partial y}(xy)-xy\frac{\partial}{\partial y}(x-y)}{(x-y)^2}=\frac{(x-y)\cdot x-xy\cdot(-1)}{(x-y)^2}=\frac{x^2-xy+xy}{(x-y)^2}=\frac{x^2}{(x-y)^2}$$
because now $x$ is treated as a constant.
Now we will evaluate these partial derivatives in the given point $(x,y)=(2,-2)$:
$$\frac{\partial f(2,-2)}{\partial x}=\left.-\frac{y^2}{(x-y)^2}\right|_{(x,y)=(2,-2)}=-\frac{(-2)^2}{(2-(-2))^2}=-\frac{4}{4^2}=-\frac{1}{4}$$
$$\frac{\partial f(2,-2)}{\partial y}=\left.\frac{x^2}{(x-y)^2}\right|_{(x,y)=(2,-2)}=\frac{2^2}{(2-(-2))^2}=\frac{4}{4^2}=\frac{1}{4}$$
