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