Answer
$$\frac{\partial f(x,y)}{\partial x}=\frac{1}{\sqrt{2x+y^3}}$$
$$\frac{\partial f(x,y)}{\partial y}=\frac{3y^2}{2\sqrt{2x+y^3}}$$
Work Step by Step
We will use chain rule to find both partial derivatives.
The partial derivative with respect to $x$ is:
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}(\sqrt{2x+y^3})=\frac{1}{2\sqrt{2x+y^3}}\frac{\partial }{\partial x}(2x+y^3)=\frac{1}{2\sqrt{2x+y^3}}\cdot2=\frac{1}{\sqrt{2x+y^3}}$$
because $y$ is treated as a constant.
The partial derivative with respect to $y$ is:
$$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(\sqrt{2x+y^3})=\frac{1}{2\sqrt{2x+y^3}}\frac{\partial }{\partial y}(2x+y^3)=\frac{1}{2\sqrt{2x+y^3}}\cdot3y^2=\frac{3y^2}{2\sqrt{2x+y^3}}$$
because now $x$ is treated as a constant.