Answer
$$\frac{\partial z}{\partial x}=2x-4y$$
$$\frac{\partial z}{\partial y}=6y-4x$$
Work Step by Step
The partial derivative with the respect to $x$ is:
$$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(x^2-4xy+3y^2)=\frac{\partial }{\partial x}(x^2)+\frac{\partial }{\partial x}(-4yx)+\frac{\partial }{\partial x}(3y^2)=\frac{\partial }{\partial x}(x^2)-4y\frac{\partial }{\partial x}(x)+0=
2x-4y\cdot1+0=2x-4y$$
because $y$ is treated as a constant.
The partial derivative with the respect to $y$ is:
$$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(x^2-4xy+3y^2)=\frac{\partial }{\partial y}(x^2)+\frac{\partial }{\partial y}(-4xy)+\frac{\partial }{\partial y}(3y^2)=0-4x\frac{\partial }{\partial y}(y)+3\frac{\partial }{\partial y}(y^2)=0-4x\cdot1+3\cdot2y=6y-4x$$
because now $x$ is treated as a constant.
