I need to prove that the jacobian of $f:\Bbb R^3 \to \Bbb R$ is constantly zero
given that $Im(f)$ (the image of $f$) is contained in $S$,
where $S$ is the points of $\Bbb R^3$ where a function $g:\Bbb R^3\to \Bbb R$ is constantly zero
given that the gradient of $g$ is never zero for all $\Bbb R^3$
I've thought of using the theorem of the implicit function, but I'm really getting nowhere. Any help will be appreciated! Thanks in advance
I will assume that $f$ is in fact a function from $\mathbb{R}^3 \to \mathbb{R}^3$ otherwise your condition doesn't make sense. You have that
$$g\circ f:\mathbb{R}^3\to\mathbb{R} $$
is constantly zero by your assumption that $im(f) \subset S$. Now take the derivative of this function at any point $x \in \mathbb{R}^3$ and use the chain rule to obtain:
$$dg_{f(x)}\circ df_x=0. $$
Now you can use that $dg_{f(x)}\neq 0$ to deduce that $df_x$ doesn't have full rank and therefore $\det(df_x)=0$.