Smoothness of division of infinitely differentiable functions Suppose I have a $C^\infty$ function $f\colon\mathbb R\to\mathbb R$, $f(0)=0$, is it true that $g(x)=\frac{f(x)}{x}$ is also a $C^\infty$?  If it is true, how do I prove it?
Generalized to multi-variable cases, suppose I have a $C^\infty$ function $f\colon\mathbb R^n\to\mathbb R$, $f(\mathbf 0)=0$, $\forall i$ $\frac{\partial f}{\partial x_i}\vert_{\mathbf x=\mathbf 0}=0$, is it true that $g_j(\mathbf x)=\frac{f(\mathbf x)}{x_j}$ are also $C^\infty$?  If it is true, how do I prove it?
 A: For smooth functions $f\colon \mathbb{R}\to\mathbb{R}$, it is the case. For a fixed $x\in \mathbb{R}$ we consider the function $h\colon [0,1] \to \mathbb{R}$ given by $h(t) = f(t\cdot x)$. Then by the fundamental theorem of calculus
$$f(x) = h(1) = h(1) - h(0) = \int_0^1 h'(t)\,dt = \int_0^1 f'(t\cdot x)\cdot x\,dt = x\underbrace{\int_0^1 f'(tx)\,dt}_{g(x)}.$$
Since $f$ is assumed to be $C^{\infty}$, one can differentiate under the integral arbitrarily often, which shows that $g$ is also $C^{\infty}$.
This does not generalise to higher-dimensional domains in the form you gave, consider for an example $f(x) = \sum_{k = 1}^n x_k^2$. On the hyperplanes $H_i := \{ x : x_i = 0\}$, the quotient isn't even defined outside the origin, we divide a nonzero real number by $0$.
It generalises however in the form that for a $C^{\infty}$ function $f\colon \mathbb{R}^n \to \mathbb{R}$ with $f(0) = 0$ there are $C^{\infty}$ functions $g_k$ such that
$$f(x) = \sum_{k = 1}^n x_k\cdot g_k(x).$$
The proof is like in the one-dimensional case,
$$f(x) = \int_0^1 \frac{\partial}{\partial t} f(tx)\,dt = \int_0^1 \sum_{k = 1}^n \frac{\partial f}{\partial x_k}(tx)\cdot x_k\,dt = \sum_{k = 1}^n x_k\int_0^1 \frac{\partial f}{\partial x_k}(tx)\,dt.$$
