show that $f$ is constant over the level sets of $\phi$ if $G = \nabla \phi $ and $fG$ is conservative

Solution 1:

The statement in your title is not quite correct. You need the level sets of $\phi$ to be connected in order for this to be true. For example, consider $f(x)=x$ and $\phi(x)=x^2$ in $\Bbb R$. Here $G=2x\vec i$ and $fG = 2x^2\vec i$ is certainly conservative. (You can easily create similar examples in higher dimensions.)

Directly, apply the chain rule. Suppose $\alpha(t)$ is any curve contained in a level set of $\phi$ and passing through $p$. Then $\phi(\alpha(t))$ is a constant function of $t$. By the chain rule, $0=\dfrac d{dt} \phi(\alpha(t)) = \nabla \phi(\alpha(t))\cdot\alpha'(t)$. Since $\nabla f$ is parallel to $\nabla\phi$ at each point, it follows that $\dfrac d{dt} f(\alpha(t)) = \nabla f(\alpha(t))\cdot\alpha'(t) = 0$ as well. This tells us that $f$ is constant along $\alpha(t)$ and hence along the connected component of the level surface passing through $p$.