Hrvatska 2011

Taking $x=0$ in the given equation we get $f(f(y))=y$ for every $y\in \mathbb{R}$. Therefore, we have $$f(y-x^2)=f(f(x^2+f(y)))=x^2+f(y).(3.1)$$ for every $x,y\in \mathbb{R}$. Taking $y=x^2$ in (3.1) we get $f(0)=x^2+f(x^2)$, that is $$f(x^2)=-x^2+f(0),$$ while taking $y=0$ in (*) gives us $$f(-x^2)=x^2+f(0).$$ Let $c=f(0)$. From the above we conclude that $$f(x)=-x+c,\forall x\in \mathbb{R}.$$ It is easy to verify that all the functions of the form $f(x)=-x+c$ for $c\in \mathbb{R}$ satisfy the given equation: $$f(x^2+f(y))=f(x^2-y+c)=-(x^2-y+c)+c=-x^2+y.$$