Answer
$$\frac{\partial^2z}{\partial x^2}=2,\frac{\partial ^2z}{\partial x\partial y}=\frac{\partial^2z}{\partial y\partial x}=-2,\frac{\partial^2z}{\partial y^2}=6$$
Work Step by Step
The partial derivative with respect to $x$ is:
$$\frac{\partial z}{\partial x}=\frac{\partial}{\partial x}(x^2-2xy+3y^2)=2x-2y$$
The partial derivative with respect to $y$ is:
$$\frac{\partial z}{\partial y}=\frac{\partial}{\partial y}(x^2-2xy+3y^2)=-2x+6y$$
The second partial derivatives are:
$$\frac{\partial ^2z}{\partial x^2}=\frac{\partial}{\partial x}\Big(\frac{\partial z}{\partial x}\Big)=\frac{\partial}{\partial x}(2x-2y)=2$$
$$\frac{\partial ^2}{\partial y^2}=\frac{\partial}{\partial y}\Big(\frac{\partial z}{\partial y}\Big)=\frac{\partial }{\partial y}(-2x+6y)=6$$
$$\frac{\partial^2z}{\partial y\partial x}=\frac{\partial}{\partial y}\Big(\frac{\partial z}{\partial x}\Big)=\frac{\partial}{\partial y}(2x-2y)=-2$$
$$\frac{\partial ^2z}{\partial x\partial y}=\frac{\partial}{\partial x}\Big(\frac{\partial z}{\partial y}\Big)=\frac{\partial }{\partial x}(-2x+6y)=-2$$
We see that the second mixed partial derivatives are equal.
