Answer
$$\frac{\partial z}{\partial x}=-\frac{y^2}{x^2}e^{y/x}$$
$$\frac{\partial z}{\partial y}=e^{y/x}+e^{y/x}\frac{y}{x}$$
Work Step by Step
The partial derivative derivative with respect to $x$ is:
$$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(ye^{y/x})=y\frac{\partial }{\partial x}(e^{y/x})=ye^{y/x}\frac{\partial }{\partial x}(\frac{y}{x})=ye^{y/x}\cdot(-\frac{y}{x^2})=-\frac{y^2}{x^2}e^{y/x}$$
because $y$ is treated as a constant. Also, we used chain rule to find $\frac{\partial }{\partial x}(e^{y/x}).$
To solve the partial derivative with respect to $y$ we will use prodact rule. The partial derivative with respect to $y$ is:
$$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(ye^{y/x})=e^{y/x}\frac{\partial }{\partial y}(y)+y\frac{\partial }{\partial y}(e^{y/x})=e^{y/x}\cdot1+y\cdot e^{y/x}\frac{\partial }{\partial y}({\frac{y}{x}})=e^{y/x}+ye^{y/x}\cdot\frac{1}{x}=e^{y/x}+e^{y/x}\frac{y}{x}$$
because here $x$ is treated as a constant. We used chain rule to find $\frac{\partial }{\partial y}(e^{y/x}).$$
