Let $f: \mathbb R \rightarrow \mathbb R$ be a diferentiable function such $f'(x) = [f(x)]^2 $ Let $f: \mathbb R \rightarrow \mathbb R$ be a differentiable function and $f(0) = 0$.
If $\forall x \in \mathbb R$ we have  $f'(x) = [f(x)]^2 $ then $f(x) = 0$,  $\forall x \in \mathbb R$
My attempt
$f$ is continuons then $f^2$ so $f'$ is continuons too.. By FTC we have $f(b) - f(a)= \int_a^b f'(t)dt$ so choosing $[0,x]$ implies $f(x) = \int_0^x [f(t)]^2dt$
and i dont know how to finish this. 
 A: Since $f'(x)=(f(x))^2$, the function $f$ is non decreasing; since $f(0)=0$, we have $f(x)\ge0$ for $x>0$. If $I=\{x>0:f(x)>0\}$ is not empty, then it is an open interval of the form $(a,\infty)$. (Note that $f(a)=0$, otherwise $a\ne\inf I$, but this is not relevant.)
Now, in the interval $I$, the function $f$ satisfies
$$
1-\frac{f'(x)}{f(x)^2}=0
$$
and so, since this is the derivative of $x\mapsto x+\frac{1}{f(x)}$,
$$
x+\frac{1}{f(x)}
$$
is constant on $I$, say $x+1/f(x)=k$. This can be written
$$
f(x)=\frac{1}{k-x}
$$
This is absurd, because $f(x)<0$ for $x>k$, contradicting the fact that $I=(a,\infty)$ and $f(x)>0$ for $x>a$.
Similarly we can exclude that $J=\{x<0:f(x)<0\}$ is non empty.
It has been observed in the comments that other contradictions can be used; for instance, the constant $k$ can be computed by taking any point $x_0>a$, so
$$
k=x_0+\frac{1}{f(x_0)}\ge x_0
$$
which means the function $f$ is not defined at $k\in(a,\infty)$: contradiction.
A: $$F(x,y)=y^2$$ is continuously differentiable, so by a theorem of Cauchy $-$ extending Picard-Lindelöf $-$ local existence and uniqueness of solutions of the differential equation $$y'=F(x,y)$$ follows. Which means, among other things, that every solution is unique where it is defined. Since $$y(x)\equiv0$$ is a solution that is defined everywhere, it is the unique solution with initial condition $$y(0)=0.$$
A: I answered a very similar one yesterday. We argue as follows. Since $f'(x) = f(x)^2$, $f(x)$ is increasing. Suppose $f(x)$ is not identically zero; we will get a contradiction. Replacing $f(x)$ by $-f(-x)$ if necessary, we can assume there is some $x_0 > 0$ for which $f(x_0) > 0$. Then since $f(x)$ is increasing, $f(x) > 0$ for all $x \geq x_0$. The condition $f'(x) = f(x)^2$ translates into
$${\partial \over \partial x}\bigg({1 \over f(x)}\bigg) = -1$$
This is equivalent to 
$${\partial \over \partial x}\bigg({1 \over f(x)} + x \bigg)= 0$$
This holds for all $x \geq x_0$. Hence for some constant $C$, for all $x \geq x_0$ one has
$${1 \over f(x)} + x = C$$
This is the same as
$$f(x) = {1 \over C - x}$$
However , for large enough $x$,  ${1 \over C - x}$ is negative, not positive as it should be since $x \geq x_0$. Thus we get a contradiction.
A: Assume that there exist $a\in\mathbb{R}$ s.t. $f(a)\not= 0$. Now since $f$ is continuous there exist an interval $I$ around $a$ where $f(x)\not = 0$ for all $x\in I$. On this interval we can safely divide the ODE by $f^2$ and integrate the equation to obtain
$$\frac{f'}{f^2} = 1\implies \int_{f(a)}^{f(x)}\frac{dy}{y^2} = \int_a^x dx \implies f(x) = \frac{f(a)}{1-(x-a)f(a)}$$
This solution shows that $f(x) \not =0$ for all $x$ and is therefore also valid for all $x$. However this contradicts $f(0) = 0$ so $f(x) \equiv 0$ is the only solution.
${\bf Edit}$: The argument above breaks down if $1-(x-a)f(a) = 0$ for $x\in(0,a)$ as we cannot integrate the ODE back to $x=0$ to give us the contradiction without integrating past a singularity. One can try to rule out this possibility, but then the other answers here give a much simpler way to arrive at the desired conclusion.
A: Your idea can work, even in a much more general setting. Since $f(0)=0$ we know that $|f(x)|\leq 1$ on some interval $[-a,a]$ where $a > 0$. Then on this interval $|f'(x)|=|f(x)|^2\leq|f(x)|$. This inequality alone is enough to ensure that on this interval $$|f(x)|\leq\frac{|x|^n}{n!}$$ for all $n\in \mathbb{N}$. Proceed by induction from the base case $n=0$: $$|f(x)|=|\int_0^xf'(t)dt|\leq\int_0^x|f'(t)|\,|dt| \leq \int_0^x|f(t)|\,|dt|\leq \int_0^{|x|}\frac{t^n}{n!}dt=\frac{|x|^{n+1}}{(n+1)!}$$
From this it follows that $f$ is indentically zero on $[-a, a]$.  This reasoning works around any zero of $f$. Therefore the set of zeroes has no supremum or infimum and so it must be the entire real line.
