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 chain rule.
The partial derivative with the respect to $x$ is:
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial}{\partial x}(\arctan\frac{y}{x})=\frac{1}{1+(\frac{y}{x})^2}\frac{\partial }{\partial x}(\frac{y}{x})=\frac{1}{\frac{x^2+y^2}{x^2}}y\frac{-1}{x^2}=-\frac{x^2}{x^2+y^2}\frac{y}{x^2}=-\frac{y}{x^2+y^2}$$
Here $y$ is treated as a constant.
The partial derivative with the respect to $y$ is:
$$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(\arctan\frac{y}{x})=\frac{1}{1+(\frac{y}{x})^2}\frac{\partial}{\partial y}(\frac{y}{x})=\frac{1}{\frac{x^2+y^2}{x^2}}\frac{1}{x}=\frac{x^2}{x^2+y^2}\frac{1}{x}=\frac{x}{x^2+y^2}$$
because now $x$ is treated as a constant.
Now we will evaluate these partial derivatives in the given point $(x,y)=(2,-2)$:
$$\frac{\partial f(2,-2)}{\partial x}=\left.-\frac{y}{x^2+y^2}\right|_{(x,y)=(2,-2)}=-\frac{-2}{2^2+(-2)^2}=\frac{2}{4+4}=\frac{2}{8}=\frac{1}{4}$$
$$\frac{\partial f(2,-2)}{\partial y}=\left.\frac{x}{x^2+y^2}\right|_{(x,y)=(2,-2)}=\frac{2}{2^2+(-2)^2}=\frac{2}{4+4}=\frac{2}{8}=\frac{1}{4}$$
