I read the following statement in a physics book:
..., $\nabla f\times \nabla p = 0$, so that $f$ can be written as $f(p)$.
Here $f,p: \mathbb{R}^3 \to \mathbb{R}$, both are smooth enough; $\nabla$ is divergence operator; $\times$ is cross product.
Is that correct? I feel it's only a necessary condition instead of sufficient.
If $ \nabla f \times \nabla p=0$ then surfaces of constant $f$ coincide with surfaces of constant $p$
Whether or not the relation between $f$ and $p$ is a functional relationship depends on some details.
Consider
$$\begin{eqnarray*} f(x,y,z)=f(x) \\p(x,y,z)=p(x) \end{eqnarray*}$$
for which we are guaranteed that $ \nabla f \times \nabla p=0$ but $f$ is a function of $p$ only if the function $p(x) $ possesses an inverse.