Answer
$$\frac{\partial z}{\partial x}=\frac{x}{y}-\frac{3y^2}{x^2}$$
$$\frac{\partial z}{\partial y}=\frac{6y}{x}-\frac{x^2}{2y^2}$$
Work Step by Step
The partial derivative with respect to $x$ is:
$$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}\left(\frac{x^2}{2y}+\frac{3y^2}{x}\right)=\frac{\partial }{\partial x}\left(\frac{x^2}{2y}\right)+\frac{\partial }{\partial x}\left(\frac{3y^2}{x}\right)=\frac{1}{2y}\frac{\partial }{\partial x}(x^2)+3y^2\frac{\partial }{\partial x}\left(\frac{1}{x}\right)=\frac{1}{2y}\cdot2x+3y^2\cdot\frac{-1}{x^2}=\frac{x}{y}-\frac{3y^2}{x^2}$$
because $y$ is treated as a constant.
The partial derivative with respect to $y$ is:
$$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}\left(\frac{x^2}{2y}+\frac{3y^2}{x}\right)=\frac{\partial }{\partial y}\left(\frac{x^2}{2y}\right)+\frac{\partial }{\partial y}\left(\frac{3y^2}{x}\right)=\frac{x^2}{2}\frac{\partial }{\partial y}\left(\frac{1}{y}\right)+\frac{3}{x}\frac{\partial }{\partial y}(y^2)=
\frac{x^2}{2}\cdot\frac{-1}{y^2}+\frac{3}{x}\cdot2y=\frac{6y}{x}-\frac{x^2}{2y^2}$$
because now $x$ is treated as a constant.