Here's a proof assuming $f$ is continuously differentiable.
Use integration by parts with $u=f(x)/x$ and $v=\frac12(x^2-x^4)$ to find
Since $f$ is convex and continuously differentiable, the function lies above all its tangents, i.e., for every $t$ and $a$ we have
f(t)\ge f(a) + f'(a)(t-a)\,.\tag2
With $t=0$ and $a=x$, and using the fact that $f(0)=0$, (2) gives
$$xf'(x)-f(x)\ge 0$$ for every $x$. The proof is complete after noting that the function $x\mapsto x^2-1$ is negative when $x>0$.