$f : \mathbb{R} \to \mathbb{R}$ differential on all $\mathbb{R}$, assume that $f(x_0) = 0$ and let $x_1 \in (x_0-\epsilon,x_0+\epsilon)$ such that $x_1 \not = x_0$ then $\int \limits_{x_0}^{x_1} \frac{1}{f(x)} dx$ diverges ?!
For example $f(x) = x^2$ so $f(0)=0$ now $\int \limits_{0}^{\epsilon} \frac{1}{x^2}dx = -\frac{1}{x}|_{0}^{\epsilon} = +\infty$
But i don't have a proof for general case ?
This is a partial solution, proving the integral is always infinite. There is a gap in the solution that assumes that there is a neighbourhood around $x_0$ where $f$ does not vanish (except on $x_0$). I don't know if this gap can be closed in the proof or if it possible to construct a function with finite integral in this gap.
We have $\lim_{h \to 0}\frac{f(x_0+h)-f(x_0)}h = \lim_{h \to 0}\frac{f(x_0+h)}h = f'(x_0) < \infty$. So there is a positive constant C and an corresponding $\epsilon > 0$ such that
$$\left|\frac{f(x_0+h)}h\right| < C, \forall |h| < \epsilon, h \neq 0$$
just choose $C=2|f'(x_0)|$ if $f'(x_0) \neq 0$ and $C=1$ otherwise.
From this follows:
$$\left|f(x_0+h)\right| \le Ch, \forall |h| < \epsilon$$
If we assume (see introductory paragraph) that $f$ has no zero (and thus does not change the sign) in $(x_0-\epsilon, x_0)$ and $(x_0, x_0+\epsilon)$, we get
$$\left|\int_{x_0}^{x_1}\frac1{f(x)}dx\right| = \int_{x_0}^{x_1}\left|\frac1{f(x)}\right|dx \ge \int_{x_0}^{x_1}\left|\frac1{C(x_1-x_0)}\right|dx = \infty$$