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\begin{align}&
\left.\begin{array}{rcl}
-\,\partiald{N}{z} & = & -3xz^{2}
\\[1mm]
\partiald{M}{z} & = & 0
\\[1mm]
\partiald{N}{x} - \partiald{M}{y} & = & z^{3}
\end{array}\right\}
\qquad\imp\qquad
\left\{\begin{array}{rcl}
N & = & xz^{3} + \fermi\pars{x,y}
\\[1mm]
M & = & \,{\rm g}\pars{x,y}
\end{array}\right.
\end{align}
Moreover,
$$
z^{3} + \partiald{f}{x} - \partiald{g}{y} = z^{3}
\qquad\imp\qquad\partiald{f}{x} = \partiald{g}{y}\tag{1}
$$
Indeed, we can always add $\ds{\nabla\Psi}$ to any $\ds{\bf F}$ we find because
$\ds{\nabla\times\pars{{\bf F} + \nabla\Psi} = \nabla\times\bf F}$. For instance, in applying the Stokes Theorem, $\ds{\oint\nabla\Psi\cdot\dd\vec{r}=0}$
The simplest choice, which satisfies $\pars{1}$, is
$\ds{\fermi\pars{x,y} = {\rm g}\pars{x,y} = 0\,,\ \forall\ x,y}$. So,
$$
M=0\,,\qquad N=xz^{3}\qquad\imp\qquad{\bf F}=xz^{3}\,{\bf j}
$$
