Behavior of derivative near the zero of a function.

There is no function $f$ that satisfies your assumptions and also satisfies $\lim \limits_{x \to 0+} \frac{f(x)}{f'(x)} \ne 0$.

Let's assume otherwise. If every interval $(0, \epsilon)$ contains a root of $f$, then the limit can only be $0$ [note that we assume that the limit exists]; Therefore we can assume that $f$ has no roots. Wlog $f > 0$ holds on $(0, \delta)$. Now we can consider $g(x) := \ln(f(x))$. Then $g$ is differentiable on $(0, \delta)$ and satisfies $\lim \limits_{x \to 0+} g(x) = -\infty$ and $\lim \limits_{x \to 0+} g'(x) = \frac{1}{L} < \infty$.

This means that on some interval $(0, \epsilon)$ the inequality $|g'(x)| \le \frac{1}{|L|} + 1$ holds. Now the mean-value theorem implies that for every $x \in (0, \epsilon)$ the inequaliy $\left|\frac{g(x) - g(\epsilon)}{x - \epsilon}\right| \le \frac{1}{|L|} + 1$ holds, i.e. $g(x) \ge -(\frac{1}{L} + 1)(\epsilon - x) + g(\epsilon)$, which is a contradiction to $\lim \limits_{x \to 0+} g(x) = -\infty$.

Note that this argumentation also works for $L = \infty$ if we set $\frac{1}{\infty} = 0$.

There is an interval $(0,\delta')$ where $f'(x) \neq 0$. Otherwise $0$ is a limit point of zeros of $f'(x)$ and the limit could not exist. By the mean value theorem, we have for each $x \in (0,\delta')$ a number $\theta$ between $0$ and $x$ such that

$$f(x) = f'(\theta)x \neq 0.$$

Hence, $f(x) > 0$ or $f(x) < 0.$ Assume WLOG that $f(x) > 0$.

Suppose

$$\lim_{x \to 0+}\frac{f(x)}{f'(x)} = L > 0.$$

There exists $0 < \delta'' < \delta'$ such that for $0 < x < \delta''$ we have

$$\frac{L}{2} < \frac{f(x)}{f'(x)} < \frac{3L}{2}.$$

Also $L,f(x) > 0 \implies f'(x) > 0$ implies $f$ is increasing.

Hence, for all $x \in (0,\delta'')$

$$\frac{f(x)}{f'(x)} = \frac{x f'(\theta)}{f'(x)} = x \frac{f(\theta)}{f(x)} \frac{f'(\theta)}{f(\theta)} \frac{f(x)}{f'(x)} < x (1)\frac{2}{L}\frac{3L}{2} = 3x \to 0.$$

Therefore, we cannot have $L>0$ and can have only $L =0$.