Differentiable function, with $f'(x)=[f(x)]^2$ Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(0)=0$ and $\forall x \in \mathbb{R}$, we have $f'(x)=[f(x)]^2$. Show that $f(x)=0, \forall x \in \mathbb{R}$.
 A: The hypotheses guarantee that $f(x)\geq0$ for all $x\geq0$. 
By contradiction, suppose that exists $a \in \mathbb{R}$ st $f(a)>0$. Since $f'=f^2\geq0$, we get that $f(x)>0$  for all $x \in (a, +\infty)$. 
Let $g: (a,+\infty) \to \mathbb{R}$ be given by  $g(x)=\frac{1}{f(x)}$. As $g'(x)=-1$ we have $g(x)=-x+c$, $c \in \mathbb{R}$. So, $f(x)=\frac1{-x+c}, \forall x \in \mathbb{R}$. From $f(a)>0$ we get that $c>a$. But then $f$ isn't continuous at $x=c$, a contradiction.
A: Let's make the Ron Gordon's argument more rigorous. Suppose that there's $x_0>0$ such that $f(x_0)\neq 0$ so by continuity of $f$ there's $\delta>0$  such that 
$$f(x)\neq 0,\quad \forall x\in(x_0-\delta,x_0]$$
and we take it maximal i.e. $f(x_0-\delta)=0$
We have
$$\int_{x_0-\alpha}^x\frac{df}{f^2}=-\frac{1}{f}\Big|_{x_0-\alpha}^x=x-x_0+\alpha,\quad\forall\alpha>\delta$$
so clearly we have a contradiction if $\alpha\to\delta$.
A: If you're not allowed to use separation of variables, then instead write your condition as ${\displaystyle {d \over dx}\bigg( {1 \over f(x)}\bigg)}= -1$ and go from there.
