So, I am trying to show the following:
Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ be a $C^{1}$ function on $\mathbb{R}^{n}$ where $f(0) = 0$. Let $g: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ be a $C^{1}$ function such that $g(x) = (g_{1}(x),g_{2}(x),...,g_{n}(x))$. Assume that $g(0) = 0$ and $Dg(0)$ is invertible. Show that there exists and open neighborhood $U$ of the origin in $\mathbb{R}^{n}$ and a continuous function $h(x) = (h_{1}(x),h_{2}(x),...,h_{n}(x)): U \rightarrow \mathbb{R}^{n}$ such that \begin{align} f(x) = g(x) \cdot h(x). \end{align} for $x \in U$.
So, right away, the inverse function theorem applies to the function $g$. However, I am not sure on how to best proceed from there. I have considered setting up a function $F(u,x) = f(x) - u(g(x))$. However, that function does not have desirable domain and range for the implicit function theorem.
Any hints or answers are welcome.
Hint: Solve this first if $g(x)=x$, and then reduce to this case by applying the Inverse Function Theorem. Do you know how to write $f(x) = \sum\limits_{i=1}^n x_ih_i(x)$?
As a further hint for the latter: Write $f(x) = \displaystyle\int_0^1 \frac d{dt} f(tx)\,dt$.