Show that $f'(x)>\frac{f(x)}{x}$ 
Let $f:[0,1]\to\mathbb{R}$ be a continuous function on $[0,1]$ and differentiable on $(0,1)$. Suppose that $f(0)=0$ and that $f'$ is strictly increasing. Show that $f'(x)>\frac{f(x)}{x}$. I would like to know if my proof holds, please and to have a feedback.

First, as $f$ is differentiable on $(0,1)$ and $f'$ is strictly increasing, then it is convex on $(0,1)$. Thus, we have the following inequality for $h<x<x+h$ ($h>0$ and such that $x+h<1$):
$\frac{f(x)-f(h)}{x-h}\le\frac{f(x+h)-f(h)}{x}\le \frac{f(x+h)-f(x)}{h}$
Applying the limit as $h\to 0$ we obtain
$\frac{f(x)-f(0)}{x}=\frac{f(x)}{x}=\lim_{h\to 0}\frac{f(x)-f(h)}{x-h}\le \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=f'(x)$ as wanted.

You proof is fine if you say that $f'$ is increasing (not $f$). However, you only get that $f'(x)\ge\frac{f(x)}{x}$, not the strict inequality.
Here is another (possibly simpler) approach which also gives the strict inequality: From the mean-value theorem we have for $x > 0$
$$
 f(x) = f(x) - f(0) = f'(c) (x-0)
$$
for some $c \in (0, x)$. Since $f'$ is strictly increasing it follows that
$$
 f(x) = f'(c) x < f'(x) x
$$
which is the desired estimate.
