I believe the following function is a counterexample:
$$f(x) = \int_0^x \frac{[\cos^2(1/t)]^{1/t^4}}{t}\, dt.$$
The idea being that the exponent $1/t^4$ is huge when $t$ is small, and should be enough to smash $\cos^2(1/t)$ down to almost $0$ so that even when divided by $t,$ we have a nice convergent improper integral. Even more, we will have $f'(0)=0.$ Note that $f'(1/(2n\pi)) = 2n\pi, n=1,2,\dots$ so that $f'$ is unbounded in every neighborhood of $0.$ A number of details remain to be checked of course ...
It's probably more instructive to play around with tall thin triangles however. So if over the points $1/n,n=1,2,\dots$ we put triangles of base lengths less than $1/n^4$ and heights $n$ (keep the bases disjoint), we arrive at a candidate integrand. Call this function $g$ and set $f(x) = x+ \int_0^x g(t)\,dt.$ This should be an example, this time with $f'(0)=1.$