Answer
$$\frac{\partial z}{\partial x}=\frac{2}{cos^2(2x-y)}$$
$$\frac{\partial z}{\partial y}=-\frac{1}{\cos^2(2x-y)}$$
Work Step by Step
We will use chain rule to find both partial derivatives.
The partial derivative with respect to $x$ is:
$$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(\tan(2x-y))=\frac{1}{\cos^2(2x-y)}\frac{\partial }{\partial x}(2x-y)=\frac{1}{\cos^2(2x-y)}\cdot2=\frac{2}{\cos^2(2x-y)}$$
because $y$ is treated as a constant.
$$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(\tan(2x-y))=\frac{1}{\cos^2(2x-y)}\frac{\partial }{\partial y}(2x-y)=\frac{1}{\cos^2(2x-y)}\cdot(-1)=-\frac{1}{\cos^2(2x-y)}$$
because now $x$ is treated as a constant.