Does little Bézout theorem hold for smooth functions?

This is actually a standard fact proven in the beginning of a manifolds course - it's a step on the way to proving all derivations are given as a linear combination of partial derivatives.

Theorem: Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ is smooth and $f(0) = 0$. Then there is a smooth function $g:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x) = xg(x)$.

Proof:

By the fundamental theorem of calculus, $\int_0^1 \frac{d}{dt}[f(tx)] dt = f(tx)|_{t=0}^{t=1} = f(x) - f(0) = f(x)$.

So, $f(x) = \int_0^1 \frac{d}{dt}[f(tx)] dt = \int_0^1 f'(tx)x dt = x\int_0^1 f'(tx) dt$.

The second equality is the chain rule ($f'$ means $\frac{d}{dx} f(x)$) and the third follows because with respect to $t$, $x$ is constant so can pull out of the integral.

Then, setting $g(x) = \int_0^1 f'(tx) dt$ gives the desired function. $\square$

To see this in action, let's suppose $f(x) = x^2 + x$. Then we see that $f'(x) = 2x + 1$ so $f'(tx) = 2tx + 1$. Thus, $g(x) = \int_0^1 2tx+1 dt = t^2x + t|_{t=0}^{t=1} = x+1$, as it should be.