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? 

Here's a proof just from basic theorems of real analysis and a little topology.
Since $f'=f^2\ge0$, $f$ is nonnegative for $x\ge0$. We have $f(x)=0$ for all $x\in[0,x^*]$; by shifting the function $x\mapsto x-x^*$ we can assume $x^*=0$, and then if $f\ne0$ we have arbitrarily small $x>0$ with $f(x)>0$. Now assume for a contradiction that $f(x)>0$ for some $x\in(0,1)$, and let $M=\min(1,f(x)/x)$. Since the set $[M,\infty)$ is closed and $f(x)/x$ is continuous with limit $0$ at $0$ (since $f'(0)=0$), there is a least $y\in(0,1)$ such that $f(y)/y\ge M$.
By the Mean Value Theorem, $f^2(z)=f'(z)=f(y)/y$ for some $0<z<y$. But if $f(y)/y\ge 1$ then $f(z)=\sqrt{f(y)/y}\ge1$, and if $1\ge f(y)/y\ge f(x)/x$ then $f(z)\ge f(y)/y$. In either case $z<y$ also has $f(z)/z\ge f(z)\ge M$, a contradiction. Thus $f=0$ on $[0,\infty)$, and by applying this to $g(x)=-f(-x)$, we have $f=0$ on $\Bbb R$.
