Answer
$$\frac{\partial z}{\partial x}=\frac{1}{2x}$$
$$\frac{\partial z}{\partial y}=\frac{1}{2y}$$
Work Step by Step
The partial derivative with respect to $x$ is:
$$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(\ln\sqrt{xy})=\frac{1}{\sqrt{xy}}\frac{\partial }{\partial x}(\sqrt{xy})=\frac{1}{\sqrt{xy}}\frac{1}{2\sqrt{xy}}\frac{\partial }{\partial x}(xy)=\frac{1}{2xy}\cdot y=\frac{1}{2x}$$
Here $y$ is treated as a constant. We used chain rule to find partial derivatives $\frac{\partial }{\partial x}(\ln\sqrt{xy})$ and $\frac{\partial }{\partial x}(\sqrt{xy})$.
The partial derivative with respect to $y$ is:
$$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(\ln\sqrt{xy})=\frac{1}{\sqrt{xy}}\frac{\partial }{\partial y}(\sqrt{xy})=\frac{1}{\sqrt{xy}}\frac{1}{2\sqrt{xy}}\frac{\partial }{\partial y}(xy)=\frac{1}{2xy}\cdot x=\frac{1}{2y}$$
Here is $x$ is treated as a constant. Again, we used chain rule to find partial derivatives $\frac{\partial }{\partial y}(\ln\sqrt{xy})$ and $\frac{\partial }{\partial y}(\sqrt{xy})$.
