If $f(0)=0$ and for every $x\in \mathbb{R}$ we have $f'(x)=[f(x)]^2$, show $f(x)=0$ $\space$ for every $x\in \mathbb{R}$ Problem
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a differentiable function such that $f(0)=0$ and for every $x\in \mathbb{R}$ we have $f'(x)=[f(x)]^2$.
Show that $f(0)=0$ $\space$ for every $x\in \mathbb{R}$.
My idea
Well, I use some EDO results to show that there is no parameter that satisfies the given initial condition $f(x)=0$, so $f$ must be identically zero.
Question
What I did makes sense?
There's some other way using just Real Analysis results to prove this? 
*This question is in the integral chapter of my book
 A: Let $U=\{x:f(x)\neq 0\}$, $U$ is open since $f$ is continuous. Let $x\in U, x=lim_nx_n, f(x_n)=0$. Since $f(x)\neq 0$, there exists an open interval $I$ containing $x$ such that for every $y\in I, f(y)\neq 0$. On $I$, we have ${{f'(x)}\over{f^2(x)}}=1$. This implies that $({1\over f})'=-1$, we deduce that on $I$, $f(t)=-{1\over{c-t}}$, there exists $N$ such that $n>N$ implies that $x_n\in I$ contradiction since ${1\over{c-x_n}}\neq 0$. We deduce that $U$ is closed and open since $U\neq\mathbb{R}$  since it does not contain $0$ and $\mathbb{R}$ is connected, we deduce that $U$ is empty.
A: Not sure I understand your idea, but it's possible to do it using real analysis.
This is a hint:
You just need to prove that $f'(x) = 0$ for all real x
If $x = 0$, $f'(x) = 0$ and let's focus on positive values of argument.
We can prove by contradiction.
Let $x_0$ be infimum of all x > 0 where $f'(x) > 0$
Just prove that at $x = x_0$ derivative does not exist. This is not hard because derivative must be zero (from the left if $x_0 > 0$ or by fact if $x_0 = 0$), but on the right we have very closed values $x > x_0$ with
$$f(x) - f(x_0) > (1 - \epsilon)(x - x_0)$$
for any small $\epsilon > 0$ we choose.
Similar thing with negative x.
