66th Czech and Slovak Mathematical Olympiad

Setting $x=1$ we learn $f(0)=f(1)y$ for all real $y$, hence $f(0)=f(1)=0$. Setting $y=1$ in the original equation we get $$f(1-x)=f(x)$$ for all real $x$. Let $t$ be an arbitrary real number. Setting $x=1-t$ gives $$f(ty)=f(1-t)y+t^{2}f(y)=f(t)y+t^{2}f(y)\text{\hspace{1em}(1)}$$ for all real $y$. Swapping the role of $t$ and $y$ we get $$f(y)t+y^{2}f(t)=f(yt)=f(ty)=f(t)y+t^{2}f(y),$$ hence $f(t)(y^2-y)=f(y)(t^2-t)$, which for $y=2$ gives $$f(t)=\frac{1}{2}f(2)(t^2-t),t\in \mathbb{R}.$$ It remains to check that the constant $a=f(2)/2$ in the derived formula $f(x)=ax(x-1)$ can be an arbitrary real number: $$\begin{array}{rl}f(x)y+(x-1{)}^{2}f(y)& =ax(x-1)y+(x-1{)}^{2}ay(y-1)\\ & =a(x-1)y(x+xy-x-y+1)\\ & =a(1-x)y((1-x)y-1)=f((1-x)y)=f(y-xy).\end{array}$$