$\nabla \varphi(f(a))\neq 0 \implies\det f'(a)=0$ Let $U\subset \mathbb R^n$ be an open subset, $f:U\to \mathbb R^n$ a differentiable function and $\varphi:\mathbb R^n\to \mathbb R$ of class $C^1$. Suppose $\varphi(f(x))=0$ for every $x\in U$.
I'm trying to prove:

If $a\in U$ and $\nabla \varphi(f(a))\neq 0$, then $\det f'(a)=0$.


Using the chain rule we know that $\varphi(f(x))'=\varphi'(f(x))\cdot f'(x)\equiv 0$ for every $x\in U$ because $\varphi\circ f$ is constant. I also know that since $\varphi\in C^1$, $\varphi'$ is continuous, thus 
$$\varphi'(f(a)+v)-\varphi'(f(a))\to 0$$ as $v\to 0$.
My problem is I don't know how to use these facts to prove this question.
 A: The thing you've written as 
$$
\varphi'(f(x))
$$
(or better, $\varphi'(f(a))$ ) can also be written 
$$
\nabla \phi (f(a)).
$$
So you have
$$
\nabla\varphi(f(a))\cdot f'(a) = 0.
$$
And you are given that $\nabla\varphi(f(a)) \ne 0$. But when you multiply that on the right by $f'(a)$, you get zero. So the transformation described by "multiply on the right by $f'(a)$" must be singular, hence the determinant of that matrix must be zero. 
General hint: when you get stuck on a problem like this, try to build a counterexample. That's what I did. I said "What if, for an open disk in the plane, for instance,  $\varphi$ is identically zero? Then the claim is false for $f(x, y) = x$." And then I looked and saw that there was a condition that $\nabla \varphi(f(a)) \ne 0$, and that led me to the answer. As an alternative: look through the hypotheses and see whether there's one you haven't used yet. In your case, that missing one was that the gradient of $\varphi$ at $a$ is nonzero. 
A: Suppose $\det (df(a)) \ne 0.$ Then $df(a)$ is nonsingular, hence surjective. Thus $df(a)[v] = \nabla \varphi (f(a))$ for some $v\in \mathbb R^n.$ Thus
$$\tag 1\langle \nabla \varphi (f(a)), df(a)[v]\rangle = \langle \nabla \varphi (f(a)), \nabla \varphi (f(a))\rangle = | \nabla \varphi (f(a))|^2 \ne 0.$$
But the left of $(1)$ is precisely $d(\varphi \circ f)(a)[v],$ and we know $d(\varphi \circ f)(a)$ is the zero linear transformation, contradiction.
