= Solution
The <curl> of the first <vector field> is
$$
\nabla\times\mathbf F=(0,-xy\sinh z,x\cosh z),
$$
which is not identically zero. Thus \b[$\mathbf F$ is not conservative] on $\mathbb R^3$ and cannot equal a negative <gradient> there.
For the second field, direct differentiation shows
$$
\mathbf G=\nabla(x^2y\cosh z).
$$
Therefore \b[$\mathbf G$ is conservative], and the requested negative-gradient potential is
$$
\boxed{\Phi=-x^2y\cosh z+C},\qquad\mathbf G=-\nabla\Phi.
$$
Its <curl> vanishes, consistently with the <curl of a gradient> identity.
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