Is this true for a continuously differentiable function Prove or find a counterexample: if $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is continuously differentiable with $f(0)=0$  then there exist continuous functions $g_1,\dots,g_n:\mathbb{R}^n\rightarrow \mathbb{R}$ with
$$f(x)=x_1g_1(x_1,\dots, x_n)+\dots+x_ng_n(x_1,\dots, x_n)$$
I don't know where to start. Should I use the implicit function theorem?
Please give me some hints.
Happy new year.
 A: Fix $x\in{\mathbb R}^n$ and consider the auxiliary function $$\phi(t):=f(t\,x)\qquad(0\leq t\leq 1)\ .$$
The chain rule then gives
$$f(x)=\phi(1)-\phi(0)=\int_0^1\phi'(t)\>dt=\int_0^1\nabla f(t\,x)\cdot x\>dt=\sum_{i=1}^n x_i\,g_i(x)$$
with
$$g_i(x):=\int_0^1 f_{.i}(t\, x)\>dt\qquad(1\leq i\leq n)\ .$$
A: For $1 \le i \le n$, for $x_i \ne 0$, define:
$$g_i(x_1,\ldots,x_n) := \frac {f(x_1, \cdots, x_{i-1}, x_i, 0, 0, \cdots, 0) - f(x_1, \cdots, x_{i-1}, 0, 0, 0, \cdots, 0)} {x_i}$$
When $x_i = 0$, let $g_i(x_1,\ldots,x_n)$ be its limit.
Note that $\displaystyle g_1(x_1,\ldots,x_n) = \frac {f(x_1, 0, \cdots, 0) - f(0, 0, \cdots, 0)} {x_1} = \frac {f(x_1, 0, \cdots, 0)} {x_1}$.
It is a trivial exercise to show that $f = \sum x_i g_i$.
The well-definedness of such functions (i.e. the fact that the limit exists) and the continuity of such functions follow from the definition of "continuously differentiable", which unfortunately you have not provided.

For example, when $n=2$, this construction amounts to:
$$f(x,y) = x \left( \frac {f(x,0)} x \right) + y \left( \frac {f(x,y) - f(x,0)} y \right)$$
When $n=3$, this construction amounts to:
$$f(x,y,z) = x \left( \frac {f(x,0,0)} x \right) + y \left( \frac {f(x,y,0) - f(x,0,0)} y \right) + z \left( \frac {f(x,y,z) - f(x,y,0)} z \right)$$
