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