Prove that $f$ is identically zero. Let $f:\Bbb{R}\to\Bbb{R}$ be differentiable such that $f(0)=0$ and holds $f'(x)=(f(x))^2$ in $\Bbb{R}$. Show that $f(x)=0$ in $\Bbb{R}$. 
What I've done:  
If $f$ is differentiable, so it is continuous. That implies that $f^2$ is continuous and, therefore, integrable. But $f$ is such that $f'=f^2$. So, by the Fundamental Theorem of Calculus, for $x>0$
$$\int _0^x(f(t))^2dt=f(x)-f(0)=f(x),$$
since $f(0)=0$. Similarly,
$$\int _{-x}^0(f(t))^2dt=f(0)-f(-x)=-f(-x).$$
This implies, for instance, that $f(x)+f(-x)=0$. But, what do I for concluding that $f(x)=0, \forall x \in \Bbb{R}$??
 A: As noted in the comments, you could invoke the standard result on the uniqueness of solutions of first order ODEs but that is a little unsatisfying. 
let $x>0.$ by Mean value theorem,., 
$$
f(x) = x f'(x_1) = x f(x_1)^2
$$
with $0 < x_1 < x. $
$$f(x_1) = x_1 f'(x_2) = x_2 f(x_2)^2$$
So we get a sequence $x_n$ tending to zero such that
$$
f(x) = x \prod \limits_{1\leq j \leq n} x_{j}^{2j} f(x_n)^{2n}
$$
Only a finite number of these terms have $x_j > 1.$
The rest tend to zero so $f(x)$ must be zero. 
Actually a little more work is needed since we must show that $x_n \to 0.$  However, they must tend to something. If it less than $1$ that is enough. So the result is true on $(-1,1)$ and $[1,1]$ by continuity... However, we can just translate the function by $1$ since the original hypothesis is translation invariant so it's true everywhere. 
A: You got that $f(x)=-f(-x)$ ($f$ is odd). 
Now :
$$f(x) = \int_0^x (f(t))^2dt = \int_0^x(-f(-t))^2dt= \int_0^x(f(-t))^2dt = f(-x)$$
$f$ is odd and is even $\implies f$ is $0$ over $\mathbb{R}$
A: Let $a\in \mathbb R$ be a point with $f(a)=0$ (for example $a=0$).
Then $|f(x)|<1$ in  an $\epsilon$-neighbourhood of $a$, where wlog. $\epsilon<1$. Let $h=\sup_{|x-a|<\epsilon}|f(x)|$, so $0\le h\le 1$. Then for $x$ in this neighbourhood
$$ |f(x)|=\left|\int_a^xf'(x)\,\mathrm dx\right| = \left|\int_a^xf^2(x)\,\mathrm dx\right|\le |x-a|h^2\le\epsilon h^2$$
$$\text{(or with MVT:)}\quad |f(x)|=|x-a|\cdot |f'(\xi)|=|x-a|f^2(\xi)\le\epsilon h^2$$ 
which implies $h\le \epsilon h^2$ and hence either $h=0$ or $1\ge h\ge \frac1\epsilon>1$. Since the latter is absurd, we have $h=0$, i.e., $f(x)=0$ in the $\epsilon$-neighbourhood. We concludethat $f^{-1}(0)$ is  a nonempty clopen subset of $\mathbb R$, so $f$ must be identically zero.
A: Look at $f'(x)f(x)=f(x)^3$. Integrating both sides:
$$f(x)^2/2=\int_0^x f'(x)f(x)dx=\int_0^x f^3(x)dx.$$
Suppose that $f(b)<0$ somewhere. Since $f$ is continuous, moving toward the origin, find the first $a$ such that $f(a)=0$. Then $f(x)<0$ on $[a,b]$ (reverse if necessary). Repeating the above integral, you get $f(b)^2/2 = \int_a^b f^3(x)dx$. But $f(x)<0$ on $[a,b]$, which leads to a contradiction. So $f(x)\geq 0$. But $f(x)$ is odd by your observation. So $f(x)=0$. 
