Answer
$f\left( {x,y,z} \right) = z{{\rm{e}}^{{x^2}}}$ is a potential function for ${\bf{F}} = \left( {2xz{{\rm{e}}^{{x^2}}},0,{{\rm{e}}^{{x^2}}}} \right)$.
Work Step by Step
Suppose that $f$ is a potential function for ${\bf{F}}$. Then, $\nabla f = {\bf{F}} = \left( {2xz{{\rm{e}}^{{x^2}}},0,{{\rm{e}}^{{x^2}}}} \right)$.
Since $\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right)$, so
(1) ${\ \ \ \ \ }$ $\left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right) = \left( {2xz{{\rm{e}}^{{x^2}}},0,{{\rm{e}}^{{x^2}}}} \right)$
There exists a function $f\left( {x,y,z} \right) = z{{\rm{e}}^{{x^2}}}$ such that equation (1) is satisfied:
$\frac{{\partial f}}{{\partial x}} = 2xz{{\rm{e}}^{{x^2}}}$, ${\ \ \ \ }$ $\frac{{\partial f}}{{\partial y}} = 0$, ${\ \ \ \ }$ $\frac{{\partial f}}{{\partial z}} = {{\rm{e}}^{{x^2}}}$
So, $f\left( {x,y,z} \right) = z{{\rm{e}}^{{x^2}}}$ is a potential function for ${\bf{F}} = \left( {2xz{{\rm{e}}^{{x^2}}},0,{{\rm{e}}^{{x^2}}}} \right)$.
