Let $f(x)$ be a differential real function defined on the real line. If $f(0)=0$ and $f'(x)=[f(x)]^2$, then $f(x)=0$ foi any $x$. Again, $f:\mathbb{R}\to\mathbb{R}$ is differentiable, $f(0)=0$, and $f'(x)=[f(x)]^2$ for every $x$. A friend suggested the following argument:
If exists $c$ such that $f(c)\neq0$, there exists an interval $I$ around $c$ such that $f(x)\neq0$ if $x\in I$ (because $f$ is continuous since it is differentiable). In that interval, we could define $g(x)=x+\frac{1}{f(x)}$. This function $g$ would be differentiable and $g'(x)=0$. Then $g(x)$ is constant, for example, $k$. Then, $f(x)=\frac{1}{k-x}$ for $x\in I$
But I don't know where to find an absurd. What should I do next?
I think I should use the fundamental theorem of calculus and try to find an absurd with $f(x)=\int_0^x f'(t)dt=\int_0^x [f(t)]^2 dt$, but I also didn't got anywhere.
Thanks in advance.

Say $f_1$ is a solution of the differential equation. Let $J \supseteq \{0\}$ be an interval of maximal length with $f_1|_{J}=0$. By continuity, it is closed. 
Assume $b=\sup J < \infty$. Then, by Picard-Lindelöf there exists an open interval containing $b$, say $\tilde J$, such that $f_1$ agrees with the 0-function on $\tilde J$, which contradicts that $J$ is maximal. That is, the assumption $b<\infty$ is wrong.
