= Solution
The condition $Df_p\ne0$ says that $0$ is a <regular value> of $f$. The <regular level set theorem> therefore makes $f^{-1}(0)$ a smooth submanifold of codimension one in $\mathbb R^3$, hence a <smooth surface>.
For
$$
F(x,y,z)=x^2+y^2-\cosh(z^2),
$$
the <gradient> is
$$
\nabla F=(2x,2y,-2z\sinh(z^2)).
$$
On $F^{-1}(0)$, $x^2+y^2=\cosh(z^2)\geq1$, so $x$ and $y$ cannot both vanish. Thus $\nabla F\ne0$ there, and the given set is a smooth surface.
Not every smooth surface in $\mathbb R^3$ is a global zero set. Every set $f^{-1}(0)$ is <closed set>[closed] because $f$ is continuous, whereas the open unit disk
$$
\{(x,y,0):x^2+y^2<1\}
$$
is a smooth surface but is not closed in $\mathbb R^3$.
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