Finding all differentiable $f: [0,+\infty) \rightarrow [0,+\infty)$ such that $f(x) = f'(x^2)$ and $f(0)=0$
As Alex already noticed, a slightly more general statement holds:
Let $f:[0, \infty) \to [0, \infty)$ be continuous, differentiable on $(0, \infty)$, and $c \ge 1 $.
If $f(0) = 0$ and $f(x) = f'(x^c)$ for all $x > 0$ then $f = 0$.
Proof: $f'(x) = f(x^{1/c}) \ge 0$, so that $f$ is increasing.
This in turn implies that $f'$ is increasing on $(0, \infty)$, so that $f$ is convex.
Step 1: $f(x) = 0$ for $0 \le x \le 1$.
From the convexity and $f(0) = 0$ it follows that $$ f(t) \le t \cdot f(1) \quad \text{ for } 0 \le t \le 1 \, . $$ On the other hand, the mean-value theorem gives $$ f(1) - f(0) = f'(\xi) (1 - 0) $$ for some $\xi \in (0, 1)$, therefore $$ f(1) = f'(\xi) = f(\xi^{1/c}) \le \xi^{1/c} \cdot f(1) \, . $$ $\xi^{1/c}$ is strictly less than one, so that $f(1) \le 0$ follows.
Since $f$ is increasing, $f(x) = 0$ for $0 \le x \le 1$.
Step 2: $f(x) = 0$ for $x \ge 1$.
For $x \ge 1$ $$ f'(x) = f(x^{1/c}) \le f(x) $$ so that we can use a standard (Grönwall's inequality type) argument: $h(x) = e^{-x} f(x)$ satisfies $$ h'(x) = e^{-x} (f'(x) - f(x)) \le 0 $$ so that $h$ is decreasing on $[1, \infty)$: $$ e^{-x} f(x) \le e^{-1} f(1) = 0 \\ \implies f(x) \le 0 \implies f(x) = 0 \, . $$
EDIT The following post was made before the condition $f(0)=0$ was stated, which leaves my critique and my counterexample inapplicable. I leave it here since I find it of interest in itself, and because if the proposed conjectures $f(x)=f(1)\cdot f_1(x)$ and $\forall x\ge 0,\;\textrm{sgn} f(x) = \textrm{sgn} f(1)$ (see below) were true, it would imply that the condition $f(0)=0$ is necessary for the conclusion $f(x)\equiv0$ to hold, and this zero function would just be a particular solution of the 'functional-differential' equation $f(x)=f'(x^2)$.
Your reasoning has a gap at the very beginning: the supremum of the set $\{x\in[0,\infty)\colon f(x)=0\}$ exists if it is both bounded above and nonempty. I think it would not be difficult to see that if it is nonempty and $f(x)\not\equiv 0$ then it will be bounded above, but I don't see why it should be nonempty anyway (unless we add the condition that $f$ be surjective).
By the way, I did some numerical approximation taking $f(1)=1$ as 'initial' condition, and I ended up with this $f$:
Here are some values.
$$\begin{array}\\x & y\\ 0.0 &0.2887337\\ 0.5 &0.5656723\\ 1.0 &1.0000000\\ 1.5 &1.5602165\\ 2.0 &2.2340116\\ 2.5 &3.0138627\\ 3.0 &3.8944997\\ 3.5 &4.8719327\\ 4.0 &5.9429892\\ 4.5 &7.1050584\\ 5.0 &8.3559366\\ \end{array}$$
The problem seems to be well conditioned and the behavior of the iterative procedure looked stable. Moreover, other functions $f$, for different initial values at $x=1$ seem to be multiples of the one given above (say $f_1$), in fact, of the form $$f(x)=f(1)\cdot f_1(x).$$ Also $f_1$ is likely positive, so the last equation and this condition would imply $$\forall x\ge 0,\;\textrm{sgn} f(x) = \textrm{sgn} f(1).$$
On the other hand, I still don't see a reasonable closed form expression for such a function.