Should $f(x) \equiv 0$ if $0\le f'(x)\le f(x)$ and $f(0)=0$?

As $f(0)=0$ and for all $x\geq0$, $f'(x)\geq0$, we have $f(x)\geq0, \forall x\geq0$.

Define $$g(x)=\mathrm e^{-x}f(x)$$ and compute $$g'(x)=\mathrm e^{-x}\left(f'(x)-f(x)\right)\leq 0$$ as $g(0)=0$ we have $g(x)\leq 0$ for all $x\geq0$. Therefore we have $$f(x)=\mathrm e^xg(x)\leq 0,\quad \forall x\geq0,$$ which proves $f\equiv0$.

To add to the very nice techniques already used, here's another way of seeing that $f = 0$. Put $M_x = \sup_{t\in[0,x]}{|f(t)|}$. (Actually, as @V.Rosetto notes, $f$ is increasing and so $M_x = f(x)$, which simplifies things!) For each $x>0$, the mean value theorem furnishes a $t\in(0,x)$ such that $f(x) = xf'(t)$. But then \begin{align*} M_x = \sup_{t\in[0,x]}|f(t)| \leq \sup_{t\in[0,x]} t|f'(t)| \leq \sup_{t\in[0,x]} t|f(t)|\leq xM_x \tag{1} \end{align*} for $x>0$. Then for $x\in(0,1)$, we must have $M_x = 0$, lest $(1)$ lead to the contradictory inequality $M_x<M_x$. Thus $f = 0$ on $[0,1)$.

But now $x\mapsto f(x-1)$ satisfies the same conditions as $f$ on $[0,\infty)$, and so it vanishes on $[0,1)$ as well. This means that $f$ vanishes on $[0,2)$. Now consider $x\mapsto f(x-2)$, and so on.

Let $x_0\in[0,x]$ be a point in which $f'$ assumes a maximum. Then

$$f(x) = f(x) - f(0) = \int_0^xf'(t)dt \le xf'(x_0) \le xf(x_0) \le xf(x)$$

so $f$ must be $0$ on $[0,1)$ and by continuity $f(1) = 0$. Since $f(x - 1)$ satisfies properties 1), 2), 3), the result follows by induction.