First note that $f$, being a derivative, has the IVP. Now suppose $f(p)\not=0$ Then $f^2(p)\not=0$ and since $f^2$ is continuous we have $f(x)\not=0$ on a nhbd of $p$. By the IVP it follows that $f(x)$ has the same sign as $f(p)$ in that nbhd. Hence $f$ itself is diff in that nbhd, since it equals $\pm \sqrt{f^2}$ with a fixed sign. Then $f=2ff'\implies f'=1/2$. Hence $f(x)=x/2+c_p$ in a nhbd of $p$. Note that this implies that $f$ is continuous on ${\bf R}$, since it is surely continuous in points where $f(p)=0$. Now suppose that $f(a)=f(b)=0$ with $a<b$. If $f(c)>0$ for some $c\in (a,b)$, then we can take $c\in [a,b]$ such that $f(c)>0$ is maximal. Then $c\in (a,b),f(c)>0$ hence $f$ is strictly increasing in a nbhd of $c$. Contradiction. Hence $f(x)\le 0$ on $[a,b]$ and also $f(x)\ge 0$ on $[a,b]$. Hence $f=0$ on $[a,b]$. It follows that $f$ is of the following type: $f(x)=x/2+c$ if $x\le a, f(x)=0$ if $a\le x\le b$ and $f(x)=x/2+d$ if $x\ge b$. Here the eventualities $a=-\infty, a=b,b=\infty$ are allowed.

i think the problem statement is vague because $f^2$ can be either $(f(x))^2$ or $f(f(x))$ .

No. It is quite common for $fg$ to mean $f(x)g(x)$ when written out.

cooljoseph wrote: No. It is quite common for $fg$ to mean $f(x)g(x)$ when written out.

$ f $ may not be differentiable. Did you think it was that easy?

Also even if $f$ was differentiable, the conclusion $2f'=1$ is not valid since you divide by $f$ which could be $0$. (Note that $f(x)=0$ is certainly also a soluiton.)