Let $f:U\rightarrow \mathbb{R}^m$ be a differentiable function over an open set $U$ and let $\phi:\mathbb{R}^m\rightarrow\mathbb{R}$ a $C^1$ function such that $\phi (f(x))=0$, $\forall x\in U$. Let $a\in U$; if $(D\phi)_b\neq 0$, where $b=f(a)$, then $det (Df)_a=0$.
I need to prove this result, and in order to do so, I defined $g=\phi \circ f$. Through the Chain Rule I reached this equation:
$ (\bigtriangledown\phi)_b (Df)_a=0$
That's true for all $a\in U$. I know I can't just say that this means that $(Df)_a=0$ so $det (Df)_a=0$. Can someone help?