Answer
$$\frac{\partial z}{\partial x}=ye^y\cos xy$$
$$\frac{\partial z}{\partial y}=e^y(\sin xy+x\cos xy)$$
Work Step by Step
The partial derivative with respect to $x$ is:
$$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(e^y\sin xy)=e^y\frac{\partial }{\partial x}(\sin xy)=e^y\cos xy\frac{\partial }{\partial x}(xy)=e^y\cos xy\cdot y=ye^y\cos xy$$
Here we treated $y$ as a constant, as we did with $e^y$. We used chain rule to find partial derivative $\frac{\partial }{\partial x}(\sin xy).$
We will use product rule to find partial derivative with respect to $y$:
$$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(e^y\sin xy)=\sin xy\frac{\partial }{\partial y}(e^y)+e^y\frac{\partial }{\partial y}(\sin xy)=\sin xy\cdot e^y+e^y\cdot \cos xy\frac{\partial }{\partial y}(xy)=e^y\sin xy+e^y\cos xy\cdot x=e^y(\sin xy+x\cos xy)$$
Here is $x$ treated as a constant. We used chain rule to find partial derivative $\frac{\partial }{\partial y}(\sin xy).$