Answer
$$\frac{\partial z}{\partial x}=-2y^2$$
$$\frac{\partial z}{\partial y}=3y^2-4xy$$
Work Step by Step
The partial derivative with the respect to $x$ is:
$$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(y^3-2xy^2-1)=\frac{\partial }{\partial x}(y^3)+\frac{\partial }{\partial x}(-2xy^2)+\frac{\partial }{\partial x}(-1)=0-2y^2\frac{\partial }{\partial x}(x)+0=-2y^2\cdot1=-2y^2$$
because $y$ is treated as a constant.
The partial derivative with the respect to $y$ is:
$$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(y^3-2xy^2-1)=\frac{\partial }{\partial y}(y^3)+\frac{\partial }{\partial y}(-2xy^2)+\frac{\partial }{\partial y}(-1)=\frac{\partial }{\partial y}(y^3)-2x\frac{\partial }{\partial y}(y^2)+0=3y^2-2x\cdot2y+0=3y^2-4xy$$
because now $x$ is treated as a constant.
