Let $F = \nabla\phi$ where $\phi \in C^2(\mathbb R^3)$ and let $f:\mathbb R^3 \to \mathbb R$ be a $C^1$ function. Prove that if $fF$ is a conservative field then $f$ is constant over the level surfaces of $\phi$
Here's what I got so far: $fF$ being a conservative field means its curl is $0$. Since $F$ is conservative itself the curl is reduced to $\nabla f \times F = \nabla f \times \nabla \phi = 0$
This last fact can be expressed by the equality $\nabla f(x) = k(x)\nabla \phi(x)$ for some scalar function $k$, if the gradient of $\phi$ never vanishes
This already intuitively tells me the answer: Since level surfaces (when smooth) are perpendicular to the gradient, and since both gradients are parallel, then the level surfaces must be the same.
Now I can actually prove the statement provided the level surfaces for $\phi$ are smooth. If the level surfaces are smooth I can parametrize any curve by a $C^1$ function $T:\mathbb R^2 \to \mathbb R^3$
Then all I need to prove is that $f \circ T$ is constant, or that $\nabla(f \circ T) = 0$
But since both $f$ and $T$ are differentiable I get
$$\nabla(f \circ T) = (\nabla f \circ T) \cdot DT = (k \circ T)(\nabla \phi \circ T) \cdot DT = (k \circ T)\nabla(\phi \circ T) = 0$$
The last step is equal to $0$ because $T$ parametrized a level curve of $\phi$ so $\phi \circ T$ is constant
The problem is that it's not always true that the level surfaces will be smooth. This is only true if $\nabla \phi$ is never $0$.
What can I do to prove this in the general case? (i.e $\nabla \phi$ is allowed to be $0$)?