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 $\sqrt \alpha > \alpha$, 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 $\alpha^2 > \alpha$. Therefore $\alpha$ is $0$ or $1$.
EDIT 2
I forgot to mention the important condition that $f(0)=0$.
 A: 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 $[1, \infty)$:
$$
 e^{-x} f(x) \le e^{-1} f(1) = 0 \\
 \implies f(x) \le 0 \implies f(x) = 0 \, .
$$
A: 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.
A: 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 $f(x) = 0$
A: My previous work was based on a false premise that $\,f(x)\,$ has a power series expansion at $\,x=0.\,$ The key insight is the correct ansatz. Since the differential equation is homogeneous, if $\,f(0) \ne 0, \,$ then WLOG assume $\, f(0)=1. \,$ Also assume that $\, f(x) \,$ has a Puiseux series expansion. Start with
 $\, f(x) = \sum_{n=0}^\infty a_n x^{e_n} ,\,$ where $\, a_0 \!=\! 1, e_0 \!=\! 0. \,$ Then 
 $\, f'(x) = \sum_{n=1}^\infty a_n e_n x^{e_n-1} ,\,$ and using the differential equation,
 $\, f(x) = \sum_{n=1}^\infty  a_n e_n x^{2e_n-2}. \,$ Comparing the two series we get the recursion
 $\, e_{n+1} = 1 + e_n/2 \,$ with solution $\, e_n = 2 - 2^{1-n}. \,$
This solution for $\, e_n \,$ leads to the recursion
 $\, a_{n+1} = a_n / e_{n+1}. \,$ Since $\, e_n \to 2 \,$ as
 $\, n \to \infty, \,$ then $\, a_n \sim 2^{-n}. \,$  This implies that
the Puiseux series for $\, f(x) \,$ converges for $\, 0 \le x \le 1. \,$
It also converges for $\, x>1 \,$ and its graph is given in the answer by Alejandro Salum. The general solution is multiplied by
 $\, f(0) \,$ and now if $\, f(0) = 0, \,$
then $\, f(x) = 0  \,$ for $\, 0 \le x. \,$ Besides this
Puiseux series expansion, it connects with the answer by Christian Blatter which shows that it also has a power series expansion at $\, x=1, \,$ coming from
 $\, g(t):=f(e^t), \, $ where $\, g(t) \,$ has a power series expansion at
 $\, t=0. \,$ As a check $\, f(0) = 1, \, f(1) \approx 3.46274661945506, \, f(2) \approx 7.73614964618559.$
Note that this question asks about a kind of delay differential equation solution. A simple example is $\,f'(x) = f(x-1)\,$ where $\,f\,$
can be defined essentially arbitrarily on $\,0\le x\le 1\,$ and then for
 $\,x>1\,$ define $\, f(x) := f(1) + \int_0^{x-1} f(t)\, dt.\,$ Of course, if $\,f(x) = 0\,$ on $\,[0,1]\,$ then $\,f(x) = 0\,$ for all $\,x\ge 0.\,$
In the current question the D.D.E. is $\, f'(x) = f(\sqrt{x}).\,$ Following the simple example, we could define $\,f\,$ arbitrarily on 
 $\,[c^{-2},c^{-1}]\,$ where $\,c>1\,$ is arbitary. The values of $\,f\,$ on  $\,(0,c^{-2})\,$ are determined by $\, f(x) = f(c^{-2}) - \int_x^{c^{-2}} f(\sqrt{t})\,dt.\,$ The values for $\,f\,$ on $\,(c^{-1},1)\,$ is more demanding. We must have $\,f\,$ infinitely differentiable on 
 $\,[c^{-2},c^{-1}]\,$ because
 $\,f(x) = f'(x^2) = 2x^2 f''(x^4) = 4x^2 f''(x^8) +8x^{10} f^{(3)}(x^8)\,$ and so on. In summary, if $\,f\,$ is infinitely differentiable on 
 $\,[c^{-2},c^{-1}]\,$ then it can be uniquely extended to $\,(0,1).\,$
