I am asking myself if for every differentiable function $f\colon[0,\infty)\to\mathbb R$ with $f(0)=0$ and $f(x)>0$ for every $x>0$ in a neighborhood of zero, we have
$$\liminf_{x\to 0^+} \frac{f(x)}{f'(x)}=0?$$
Consider $\displaystyle\lim_{\delta\to 0+} \inf\Big\{\frac{f(x)}{f'(x)}:x\in (0,\delta)\Big\}$.
As the function $f$ is continuous, the numerator goes to $0$. Since $f$ is continuous at $0$ and $f(x)>0\ \forall x>0$, we have by the $\varepsilon,\delta$-criterion: $$\forall \varepsilon >0 \ \exists \delta>0 \text{ such that } \forall x \in [0,\delta) \text{ we have } |f(x)|=f(x)<\varepsilon.$$
Hence, the difference quotient $\frac{f(x)- f(0)}{x-0}=\frac{f(x)}{x}>0 \ \forall x>0$. My thought is to use the following intuitive argument: If $f(0)=0$ and $f(x)>0$ for every $x>0$, then $f'(x)>0$ for $x$ sufficiently close to $0$.
The function $$ g(x) = \begin{cases} x^2\sin(\frac{1}{x}) + \frac{x}{10}, & x > 0 \\ 0, & x = 0 \end{cases} $$ is a counterexample. You can check that $g(x)>0$ for all $x > 0$ sufficiently small, and that $g$ is differentiable on $[0,\infty)$.
First I will give a heuristic argument, and then I will sketch a more detailed one. Indeed, choose a sequence $x_n\searrow 0$ such that $g'(x_n)>-x_n$ (say by looking very closely to the right of a sequence of local maxima of $g$). For $n$ sufficiently large, $g(x_n) < x_n$ as the quadratic term becomes negligible, so for these points $x_n$, $$ \frac{g(x_n)}{g'(x_n)} <\frac{x_n}{-x_n} = -1. $$ Hence $\liminf_{x\to 0^+}\frac{g(x)}{g'(x)} \le -1$.
Being a little bit more careful, I think it is possible to prove the following lemma. Let $\{y_n\}$ be the sequence of local maxima of $g$ in the interval $(0,\frac{1}{10})$, arranged so that $y_1 > y_2 > \dots$.
Lemma. There exists constants $C>c>0$ so that for each $n=1,2,\dots$, $-C < g''(y_n) < -c$.
The idea of the lemma is that near a local maximum of $g$, $g$ looks like an upside-down parabola with a fixed aspect ratio (the parabolas are not getting relatively skinnier or fatter as $x\to 0^+$).
Next, expand $g'(y_n+\epsilon_n) \approx g''(y_n)\epsilon_n \approx -\epsilon_n$, and set $x_n = y_n + \epsilon_n$. For $n$ very large, we still have $g(x_n) < x_n$ as the quadratic term becomes negligible, and we also have $$ \frac{g(x_n)}{g'(x_n)} \approx \frac{x_n}{-\epsilon_n}. $$ Note the freedom we have to choose $\epsilon_n$ as small as we like, say $\epsilon_n < x_n^2$. This shows how we can prove for this function $g$, we have $$ \liminf_{x\to 0^+}\frac{g(x)}{g'(x)} = -\infty $$