Answer
$\phi(x,y,z)=\dfrac{x^4}{4}+y^2-\dfrac{z^4}{4}$
Work Step by Step
For a vector field to be Conservative, $\dfrac{\partial f}{\partial y}=\dfrac{\partial g}{\partial x}$ and $\dfrac{\partial g}{\partial z}=\dfrac{\partial h}{\partial y}$ and $\dfrac{\partial f}{\partial z}=\dfrac{\partial h}{\partial x}$
We have: $f(x,y,z)=x^3, g(x,y,z) =2y$ and $h(x,y,z)=-z^3$
Thus, $\dfrac{\partial f}{\partial y}=0=\dfrac{\partial g}{\partial x}$ and $\dfrac{\partial g}{\partial z}=0=\dfrac{\partial h}{\partial y}$ and $\dfrac{\partial f}{\partial z}=0=\dfrac{\partial h}{\partial x}$
Therefore, a vector field $F$ is Conservative.
Now, potential function $F=\nabla \phi$
So, $\dfrac{\partial \phi}{\partial x}=x^3 \implies \phi(x,y,z)=\dfrac{x^4}{4}+a(y,z)$
and $\dfrac{\partial \phi}{\partial y}=2y=\dfrac{\partial a}{\partial y} \implies a(y,z)=y^2+b(z)$ and $\dfrac{\partial \phi}{\partial z}=-z^3 =b'(z) \implies b(z)=\dfrac{-z^4}{4}$
Thus, $\phi(x,y,z)=\dfrac{x^4}{4}+y^2-\dfrac{z^4}{4}$
