Selection tests for the Balkan Mathematical Olympiad 2013

Plug in $y=0$. The functional equation becomes $$f(f(f(x)))=x+f(0),$$ for all $x\in \mathbb{R}$. Since the map $x\mapsto x+f(0)$ is bijective, then so is $f$.

Plug in $y=-x$. The functional equation becomes $$f(f(f(x)-x)-x)=f(-x),$$ for all $x\in \mathbb{R}$. By injectivity of $f$, we can cancel $f$ in both sides and obtain $$f(f(x)-x)=0,$$ for all $x\in \mathbb{R}$. By surjectivity of $f$ there exists a real number $a$ such that $f(a)=0$. Again by cancelling $f$ from both sides we obtain $$f(x)=x+a$$ for all $x\in \mathbb{R}$. But $0=f(a)=2a$. We deduce that $$f(x)=x$$ for all $x\in \mathbb{R}$. Conversely, we check easily that this function is a solution to the problem.