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