Answer
$$\frac{\partial^2z}{\partial x^2}=12x^2-6y^2,\frac{\partial ^2z}{\partial x\partial y}=\frac{\partial^2z}{\partial y\partial x}=-12xy,\frac{\partial^2z}{\partial y^2}=12y^2-6x^2$$
Work Step by Step
The partial derivative with respect to $x$ is:
$$\frac{\partial z}{\partial x}=\frac{\partial}{\partial x}(x^4-3x^2y^2+y^4)=4x^3-3y^2\cdot2x=4x^3-6xy^2$$
The partial derivative with respect to $y$ is:
$$\frac{\partial z}{\partial y}=\frac{\partial}{\partial y}(x^4-3x^2y^2+y^4)=-3x^2\cdot2y+4y^3=-6x^2y+4y^3$$
The second partial derivatives are:
$$\frac{\partial ^2z}{\partial x^2}=\frac{\partial}{\partial x}\Big(\frac{\partial z}{\partial x}\Big)=\frac{\partial}{\partial x}(4x^3-6xy^2)=12x^2-6y^2$$
$$\frac{\partial ^2}{\partial y^2}=\frac{\partial}{\partial y}\Big(\frac{\partial z}{\partial y}\Big)=\frac{\partial }{\partial y}(-6x^2y+4y^3)=-6x^2+12y^2$$
$$\frac{\partial^2z}{\partial y\partial x}=\frac{\partial}{\partial y}\Big(\frac{\partial z}{\partial x}\Big)=\frac{\partial}{\partial y}(4x^3-6xy^2)=-12xy$$
$$\frac{\partial ^2z}{\partial x\partial y}=\frac{\partial}{\partial x}\Big(\frac{\partial z}{\partial y}\Big)=\frac{\partial }{\partial x}(-6x^2y+4y^3)=-12xy$$
We see that the second mixed partial derivatives are equal.
