= Solution
For a continuously differentiable <vector field> on all of $\mathbb R^3$, the necessary and sufficient condition for a <conservative vector field> is \b[$\nabla\times\mathbf F=0$]. Necessity follows from equality of <mixed partial derivatives> of a potential; sufficiency uses the $\mathbb R^3$ being a <simply connected space>. On a general domain the <topology> cannot be omitted.
Here the relevant <mixed partial derivatives> are
$$
\partial_yF_z=\partial_zF_y=2ye^z,\quad
\partial_zF_x=\partial_xF_z=-6z^2,\quad
\partial_xF_y=\partial_yF_x=-2x\sin y.
$$
Thus the <curl> vanishes. Integrating the first component in $x$ gives $\Phi=x^2\cos y-2xz^3+A(y,z)$. Matching the second component gives $A_y=3+2ye^z$, hence $A=3y+y^2e^z+B(z)$. Matching the last component forces $B'=0$. A <potential of a conservative vector field> with the convention $\mathbf F=\nabla\Phi$ is consequently
$$
\boxed{\Phi=x^2\cos y-2xz^3+3y+y^2e^z+C.}
$$
If a physical potential is defined instead through $\mathbf F=-\nabla V$, it is $V=-\Phi$.
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