If $f: \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function, $f(0)=0$ and $f' = f^2$, then $f = 0$. If $f: \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function, $f(0)=0$ and $f' = f^2$, then $f = 0$. Any help? 

Note that your function is increasing due to $f' = f^2 \geq 0$. So if you have such a function which is not identically zero, the set of $x$ for which $f(x) = 0$ is some closed interval $I$ (possibly unbounded) which is not all of ${\mathbb R}$. Let $x_0$ be an endpoint of $I$. Replacing $f(x)$ by $-f(-x)$ if necessary, we can assume $x_0$ is the right endpoint of $I$.
On $(x_0,\infty)$ the equation $f' = f^2$ is separable, and standard calculus techniques give that $f(x) = -{1 \over x + c}$ for $x > x_0$ for some constant $c$. But then taking limits as $x \rightarrow x_0$ gives $f(x) = -{1 \over x_0 + c}$, contradicting that $f(x_0) = 0$ since $x_0 \in I$. 
This contradiction shows that $f = 0$ is the only solution.
By the way, solving $f' = f^2$ here only uses basic real analysis techniques... it's the same as ${d \over dx} ({1 \over f(x)} + x) = 0$, which by the mean value theorem for example says that ${1 \over f(x)} + x = c$ for some constant $c$. This implies that $f(x) = -{1 \over x + c}$.
