Answer
$$\frac{\partial f(x,y)}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}$$
$$\frac{\partial f(x,y)}{\partial y}=\frac{y}{\sqrt{x^2+y^2}}$$
Work Step by Step
To solve both partial derivatives we will use chain rule.
The partial derivative with respect to $x$ is:
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}(\sqrt{x^2+y^2})=\frac{1}{2\sqrt{x^2+y^2}}\frac{\partial }{\partial x}(x^2+y^2)=\frac{1}{2\sqrt{x^2+y^2}}\cdot2x=\frac{x}{\sqrt{x^2+y^2}}$$
because $y$ is treated as a constant.
The partial derivative with respect to $y$ is:
$$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(\sqrt{x^2+
y^2})=\frac{1}{2\sqrt{x^2+y^2}}\frac{\partial }{\partial y}(x^2+y^2)=\frac{1}{2\sqrt{x^2+y^2}}\cdot2y=\frac{y}{\sqrt{x^2+y^2}}$$
because now $x$ is treated as a constant.