Suppose the function $f:(0,1] \to \mathbb{R}$ is differentiable and satisfies $f(x) \geqslant 0$ and $\lim_{x \to 0+}f(x) = 0$.
One way demonstrate that the derivative of a function tends to behave worse than the function is as follows. Show that the logarithmic derivative $f'/f$ diverges to $\pm \infty$ as $ x \to 0+$ or is, at least, unbounded in a neighborhood of $x =0$.
This is quite simple to show under more restrictive conditions. If we assume that that $f'/f$ is positive, bounded and integrable on $[\delta,1]$ for all $\delta > 0$, then, with a change of variable $y = f(x),$ we have
$$\lim_{ \delta \to 0+}\int_{\delta}^{1}\frac{f'(x)}{f(x)}\,dx = \lim_{ \delta \to 0+}\int_{f(\delta)}^{f(1)} \frac{dy}{y} = \lim_{ \delta \to 0+}[\log f(1) - \log f(\delta)] = + \infty,$$
and this implies that $f'(x)/f(x) \to +\infty$ as $x \to 0+.$
The problem then is to show that, under the weaker conditions where $f'/f$ need not be integrable, etc., we have either divergence or unboundedness of $f'/f$ near $x =0$. I suspect this is true, but I am not sure that a divergent limit is always necessary.
I think the question can be resolved by applying the mean value theorem. For any points $y > x >0$ there exists $\xi_{x,y}$ between $x$ and $y$ such that
$$\log f(y) - \log f(x) = \frac{f'(\xi_{x,y})}{f(\xi_{x,y})}(y-x).$$
Thus,
$$\lim_{x \to 0+} \frac{f'(\xi_{x,y})}{f(\xi_{x,y})} = [\log f(y) - \lim_{x \to 0+} \log f(x)]/y = + \infty.$$
Hence, on any interval $(0,y]$ no matter how small we can find a sequence of points $(x_n)$ such that $f'(x_n)/f(x_n) \to + \infty.$ The mean value theorem is non-constructive with respect to the intermediate point, so we cannot determine that $x_n \to 0$ or more generally that $\lim_{x \to 0+} \xi_{x,y}= 0.$ This shows, at least, that $f'/f$ must be unbounded in any neighborhood of $x=0.$
An example would be $f(x) = x^2\sin(1/x),$ where
$$\frac{f'(x)}{f(x)} = \frac{2}{x}- \frac{\cot(1/x)}{x^2}.$$
Here the limit does not exist, but the logarithmic derivative is clearly unbounded.