Answer
$$\frac{\partial z}{\partial x}=\frac{y^2}{\sqrt x}$$
$$\frac{\partial z}{\partial y}=4y\sqrt x$$
Work Step by Step
The partial derivative with the respect to $x$ is:
$$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(2y^2\sqrt x)=2y^2\frac{\partial }{\partial x}(\sqrt x)=2y^2\cdot\frac{1}{2\sqrt x}=\frac{y^2}{\sqrt x}$$
because $y$ is treated as a constant.
The partial derivative with the respect to $y$ is:
$$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(2y^2\sqrt x)=2\sqrt x\frac{\partial }{\partial y}(y^2)=2\sqrt x2y=4y\sqrt x$$
because now $x$ is treated as a constant.