$g(x) = \frac{f'(x)}{f(x)}$ not bounded

Hint: observe that $g(x) = h'(x)$, with $h(x) := \log f(x)$, and that, by assumption, $h(x) \to -\infty$ as $x\to 0^+$. Assume by contradiction that $h'$ is bounded. Can $h(x)\to -\infty$ as $x\to 0^+$?

By the mean value theorem, for any $0 < x < y < 1$, there exists $\xi(x,y) \in (x,y)$ such that

$$\frac{\log f(y) - \log f(x)}{y-x} = \frac{f'(\xi(x,y))}{f(\xi(x,y))}= g(\xi(x,y))$$

Thus,

$$\lim_{x \to 0+}g(\xi(x,y)) = \lim_{x \to 0+}\frac{\log f(y) - \log f(x)}{y-x}= \frac{\log f(y) - \lim_{x \to 0+}\log f(x)}{y} = +\infty$$

Hence, for any $n \in \mathbb{N}$ there exists $\delta_n \in (0,y)$ such that $g(\xi(\delta_n,y))> n$.

As the mean value theorem is non-constructive it is not possible to determine if $\xi(\delta_n,y) \to 0$ with $y$ fixed as $n \to \infty$. However, since $y$ can be chosen arbitrarily close to $0$, this shows that $g$ is unbounded in every neighborhood of $0$.