Finding all $f:[0, \infty) \to [0, \infty)$ differentiable and convex with $f(0)=0$ and $f'(x)\cdot f\bigl(f(x)\bigr)=x$ 
Find all functions $f:[0, \infty) \to [0, \infty)$, differentiable and convex, such that
$$f(0)=0 \tag1\label1$$
and
$$ \ f'(x)\cdot f\bigl(f(x)\bigr)=x, \forall x \tag2\label2$$


Obviously, $f(x)=x$ is a solution, so I'm trying to find other solutions.
From \eqref{2} we get $f(x) \gt 0, \ f'(x) \gt 0, \forall x \gt 0$ and $f'(0)=0$ therefore $f$ is strictly increasing.
So far, I don't know how to use the convexity of $f$, the definition of convexity doesn't seem to help.
UPDATE:
From \eqref{2} $f'(x)=\dfrac x{f\bigl(f(x)\bigr)}, \ \forall x \gt 0$ therefore $f$ is twice differentiable on $(0, \infty)$.
 A: The only function that is convex, has $f(0)=0$ and satisfy $f'(x)f(f(x)) = x$ is the identity $f(x) = x$.

Since $f$ is convex and $f(0) = 0$ we have $$f(xt) = f(0\cdot (1-t) + x\cdot t) \leq (1-t)f(0) + t f(x) = tf(x) \implies \frac{f(xt)}{xt} \leq \frac{f(x)}{x}$$ for all $t\in[0,1]$ and $x\in\mathbb{R}_{\geq 0}$. This shows that $g(x) = \frac{f(x)}{x}$ is an increasing function. 
We now either have $g(x) < 1$ for all $x$, $g(x) > 1$ for all $x$ or there exist a $x_0$ such that $g(x_0) = 1$. If $g(x) < 1$ for all $x$ we define $x_0 = \infty$ and if $g(x) > 1$ for all $x$ we define $x_0 = 0$.

On $[0,x_0)$ we have $f(x) \leq x$  so 
$$g(f(x)) \leq g(x) \implies f(f(x)) \leq \frac{f^2(x)}{x}$$ and the functional equation gives $$x \leq f'(x) \frac{f^2(x)}{x} \implies 0 \leq \frac{d}{dx}[f^3(x) - x^3] \implies x \leq f(x)$$ and since $f(x) \leq x$ and $x\leq f(x)$ we have $f(x) = x$ on $[0,x_0)$.

On $(x_0,\infty)$ we have $f(x) \geq x$ so 
$$g(f(x)) \geq g(x) \implies f(f(x)) \geq \frac{f^2(x)}{x}$$
and the functional equation gives $$x \geq f'(x) \frac{f^2(x)}{x} \implies 0 \geq \frac{d}{dx}[f^3(x) - x^3] \implies x\geq f(x)$$ and again since $f(x)\geq x$ and $f(x) \leq x$ we have $f(x) = x$ on $(x_0,\infty)$. It follows that $f(x) = x$ for all $x$.

Some more details on the integration of the inequality above. Since $0 \leq \frac{d}{dx}[f^3(x) - x^3]$ on $[0,x_0)$ we have
$$0 =\int_0^x 0{\rm d}x \leq \int_0^x \frac{d}{dx}[f^3(x)-x^3)]{\rm d}x = f^3(x) - f^3(0) - x^3 + 0^3 = f^3(x) - x^3 \\\implies x^3 \leq f^3(x) \implies x \leq f(x)$$
for $0\leq x < x_0$ and since $0 \geq \frac{d}{dx}[f^3(x) - x^3]$ on $(x_0,\infty$) we have
$$ 0 =\int_{x_0}^x 0{\rm d}x \geq \int_{x_0}^x \frac{d}{dx}[f^3(x)-x^3)]{\rm d}x = f^3(x) - f^3(x_0) - x^3 + x_0^3 = f^3(x) - x^3\\f^3(x) \leq x^3 \implies f(x) \leq x$$ 
for $x_0 < x$ where we have used $f(x_0) = x_0$ to simplify.
