Mathematica competitions in Croatia

Letting $x=y=-\alpha$ we get $$f(f(-\alpha ))=f(f(-\alpha ))+f(\alpha )-\alpha ⟹f(\alpha )=\alpha .$$ Letting $x=-\alpha ,y=\alpha$ we get $$\alpha =f(f(\alpha ))=f(f(-\alpha ))+\alpha +\alpha ⟹f(f(-\alpha ))=-\alpha .$$ Now let us denote $f(-\alpha )=a$. We know that $f(a)=-\alpha$. Letting $x=\alpha ,y=a$ we get $$\alpha =2\alpha +a⟹a=-\alpha .$$ Finally, letting $y=-\alpha$ gives us $$f(x)=f(f(x)),\forall x\in \mathbb{R},$$ and letting $x=-\alpha$ gives us $$f(f(y))=y,\forall y\in \mathbb{R}.$$ It follows that $$f(x)=f(f(x))=x,\forall x\in \mathbb{R}.$$ It is easy to check that the function $f(x)=x$ really is a solution.