If $f(0)=0$ and $f'$ is increasing, then $\frac{f(x)}{x}$ is increasing.

$${f(x)\over x}=\int_0^1 f'(t\, x)\ dt\qquad(x>0)\ .$$

Note that $g$ is differentiable everywhere it is defined. Fix $x, y\in\mathbb{R}$ with $y>x$. By the mean-value theorem, for some $t\in(x,y)$, we have

$$\frac{g(y)-g(x)}{y-x} = g'(t)= \frac{tf'(t)-f(t)}{t^2}.$$

Because $y-x$ is positive, it suffices to show that $$h(x)=\frac{xf'(x)-f(x)}{x^2}$$ is nonnegative. This reduces to showing that

$$f'(x)\ge \frac{f(x)}{x}$$

for $x>0$.

Consider the mean-value theorem applied to $[0,x]$ For some $t\in(0,x)$,

$$\frac{f(x)-f(0)}{x}=f'(t),$$

and the result follows because $f(0)=0$ and $f'$ is monotonically increasing.