If $f(x)$ is smooth and odd, must $f(x)/x$ be smooth?

$g$ will always be well defined at $0$ because:

$$g(0) = \lim_{x\to 0} \frac{f(x)}{x} = \lim_{x\to 0} \frac{f(x)-0}{x-0} \equiv f'(0)$$

by the definition of the derivative. One can even prove via repeated use of quotient rule and induction that by this definition,

$$g^{(n)}(0) = \frac{f^{(n+1)}(0)}{(n+1)}$$

which is well defined because $f$ is smooth.

Assume that $f:\mathbb{R}\to\mathbb{R}$ is smooth and $f(0) = 0$. If we define $g$ as in OP's construction, then $g$ is well-defined and the following representation holds for all $x \in \mathbb{R}$.

$$ g(x) = \int_{0}^{1} f'(xt) \, \mathrm{d}t. $$

Then it is easy to prove that Leibniz integral rule is applicable here, showing that $g$ is indefinitely differentiable and

$$ g^{(n)}(x) = \int_{0}^{1} \frac{\partial^n}{\partial x^n} f'(xt) \, \mathrm{d}t = \int_{0}^{1} t^n f^{(n+1)}(xt) \, \mathrm{d}t. $$

In particular, we immediately find that $g^{(n)}(0) = \int_{0}^{1} t^n f^{(n+1)}(0) \, \mathrm{d}t = \frac{f^{(n+1)}(0)}{n+1}$.

I accepted Ninad Munshi's answer because it's correct and led me to the full proof, but I will put the full proof here for posterity:

Combining the formulas $$\left( \frac{d}{dx}\right)^k\left(\frac{1}{x}\right)=\frac{(-1)^k k!}{x^{k+1}}$$ and $$\left( \frac{d}{dx}\right)^n \big( p(x)q(g) \big)=\sum_{k=0}^n \frac{n!}{k!(n-k)!}p^{(k)}(x)q^{(n-k)}(x),$$

We have, for $x\neq 0$:

$\begin{align} \\g^{(n)}(x) &\equiv \left( \frac{d}{dx}\right)^n \left( \frac{1}{x}\cdot f(x)\right) \\ \\ \\ &= \sum_{k=0}^n \frac{n!}{k!(n-k)!}\cdot \frac{(-1)^k k!}{x^{k+1}} f^{(n-k)}(x) \\\\\\ &= \frac{1}{x^{n+1}}\sum_{k=0}^n \frac{n!}{(n-k)!}\cdot (-1)^k \cdot x^{n-k} f^{(n-k)}(x).\\ \\ \end{align}$

Now assume the formula $g^{(n)}(0)=\frac{f^{(n+1)}(0)}{n+1}$ holds for some $n$, and we hope to then show that it holds for $n+1$. By definition of the derivative, we have:

$\begin{align} \\ g^{(n+1)}(0) &= \lim_{x\to 0}\frac{g^{(n)}(x)-g^{(n)}(0)}{x-0} \\\\\\ &= \lim_{x\to 0}\frac{\left(\sum_{k=0}^n \frac{n!}{(n-k)!}\cdot (-1)^k \cdot x^{n-k} f^{(n-k)}(x) \right)-x^{n+1}\frac{f^{(n+1)}(0)}{n+1}}{x^{n+2}} \\\\ \end{align}$

The only term in the numerator without a factor of $x$ in front is the term with $f^{(0)}(x)$, which must also vanish as $x\to 0$ because $f$ is odd, so we are justified in using L'Hopital's rule. When we do, the sum telescopes, leaving only:

$\begin{align} \\ g^{(n+1)}(0) &= \lim_{x\to 0}\frac{x^n f^{(n+1)}(x)-x^n f^{(n+1)}(0)}{(n+2)x^{n+1}} \\\\\\ &= \lim_{x\to 0}\frac{f^{(n+1)}(x)-f^{(n+1)}(0)}{(n+2)x}\\\\ \end{align}$

And then one more application of L'Hopital's rule shows the formula holds for $n+1$. Ninad's post already clearly shows the formula indeed holds for $n=0$, so we are done. Interestingly, we only needed to assume that $f$ is smooth and $f(0)=0$, not that $f$ is odd.