Finding all differentiable $f: [0,+\infty) \rightarrow [0,+\infty)$ such that $f(x) = f'(x^2)$ and $f(0)=0$ After some investigation it seems fairly obvious to me that the only such function is the zero function, however I haven't been able to prove it. By considering $$\alpha  =\sup\{x\in[0,+\infty) :f(x) = 0\},$$ I was able to show that $\alpha$ can only be $1$ or $0$ but I could not weed out those two possibilities. Any hints/solutions welcome.

EDIT 1 

Because of the continuity of $f$, we must have $f(\alpha) = 0$. Note that because of the relation given we have $$\int_0^{\sqrt \alpha}2xf'(x^2)\,\mathrm dx = f(\alpha),$$ but because of the relationship given this implies
$$\int_0^{\sqrt \alpha}2xf(x)\,\mathrm dx = f(\alpha).$$
If $\alpha$ is strictly between $0$ and $1$, then $\sqrt \alpha &gt; \alpha$, but then splitting the integral we get 
$$\int_{\alpha}^{\sqrt \alpha}2xf(x)\,\mathrm dx = f(\alpha) = 0.$$ But by our choice of $α$, this integral should be non-zero since our function is positive. Hence $\alpha$ cannot be between $0$ and $1$.

Now suppose it is greater than $1$, then we have $$f(\alpha^2) =\int_0^{\alpha}2xf(x)\,\mathrm dx = 0.$$
Since our function is $0$ on $[0,\alpha]$ (Note that it is increasing), this is again a contradiction because $\alpha^2 &gt; \alpha$. Therefore $\alpha$ is $0$ or $1$.

EDIT 2

I forgot to mention the important condition that $f(0)=0$.
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 $x &gt; 0$ 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 $[1, \infty)$:
$$
 e^{-x} f(x) \le e^{-1} f(1) = 0 \\
 \implies f(x) \le 0 \implies f(x) = 0 \, .
$$
You can use chain rule to get:

$f'(x) = 2xf''(x^2)$

Now you have:

$f'(x^2) = 2 x^2 f''(x^4)$

Now substitute it on first equation:

$f(x) = 2 x^2 f''(x^4)$

And since $x \in [0,\infty)$ we have 3 cases:

(i) f(x) is constant $\Rightarrow$ $f(x) = f^{(n)}(x)=0$

(ii) f(x)  is non decreasing $\Rightarrow$ $f^{(n)}(x)\geq0$, but in order to equations to hold $f^{(n)}(x)=0$; $n=0,1,...$

(iii)f(x)  is non increasing $\Rightarrow$ $f^{(n)}(x)\leq0$, but in order to equations to hold $f^{(n)}(x)=0$; $n=0,1,...$

Concluding that $f(x) = 0$