Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ of class $c^\infty$ such that the only point where it's zero is at the origin.
I want to show that $\nabla(0,0) = (0,0)$.
Looking for a contradiction, I'll start assuming that $f_x(0,0) \neq 0$. Then we can apply the Implicit Function Theorem.
If $S = \lbrace (x,y): f(x,y) = 0 \rbrace = \lbrace (0,0) \rbrace$, by the theorem we know that there are:
- An open interval $U \subset \mathbb{R}$ around $0$
- An open interval $V \subset \mathbb{R}$ around $0$
- A $c^1$ function from $U$ to $V$ such that its graph is $S \cap V$ .
Is the fact that the theorem talks about "open" intervals be enough to assert a contradiction?
EDIT: I was also thinking that since $\phi(x)$ is $c^1$ it should map open sets to open sets, but the graph has a single point, so that is the contradiction.