Saudi Arabian IMO Booklet

Put $x=0$, we get $2f(0)f(y)-f(0)=0$, if $f(0)\ne 0$ then $f(y)=1/2$ for all $y$, which does not satisfy. Thus $f(0)=0$.

Put $y=0$, we get $2f(x{)}^2-f(x^2)=x/2\cdot f(2x)$ then plugging back $2f(x)\cdot f(x+y)=2f(x{)}^2+2x\cdot f(f(y))$ so $$f(x)\cdot f(x+y)=f(x{)}^2+x\cdot f(f(y)).(\diamond )$$ Note that $f(-x{)}^2=x\cdot f(f(x))$. In $(\diamond )$, take $y=-2x$ then $$f(x)\cdot f(-x)=f(x{)}^2+x\cdot f(f(-2x)).$$ Note that $-2x\cdot f(f(-2x))=f(2x{)}^2$, so $x\cdot f(f(-2x))=-f(2x{)}^2/2$. Thus $$f(x)\cdot f(-x)=f(x{)}^2-f(2x{)}^2/2$$ or $$-2f(x)f(-x)+2f(x{)}^2=f(2x{)}^2.$$ In $(\diamond )$, take $x=y\ne 0$ then $$f(x)\cdot f(2x)=f(x{)}^2+x\cdot f(f(x))$$ or $$f(2x)=f(x)+x\cdot f(f(x))/f(x)=f(x)+f(-x{)}^2/f(x).$$ So $(f(x)+f(-x{)}^2/f(x){)}^2=2f(x{)}^2-2f(x)\cdot f(-x)$. Dividing both sides by $f(-x{)}^2$ and put $k=f(x)/f(-x)$ then $$(k+1/k{)}^2=2k^2-2k⟷k^2-2k=2+1/k^2.$$ Changing the sign $x\to -x$ then we get $(k+1/k{)}^2=2/k^2-2/k$ combining with above, we get $k=-1$. Then $f(x)$ is odd. Now we get $f(x{)}^2=x\cdot f(f(x))$. Put $y\to -y$, we get $$f(x)\cdot f(x-y)=f(x{)}^2+x\cdot f(f(-y)),$$ $$f(x)[f(x+y)+f(x-y)]=2f(x{)}^2.$$ Consider $x\ne 0$ then $f(x+y)+f(x-y)=2f(x)$ using $f(0)=0$, this implies that $f(x)$ is additive. Note that $$2f(x{)}^2-f(x^2)=x/2\cdot f(2x)=x\cdot f(x)⟺2f(x{)}^2=f(x^2)+xf(x).$$ Plugging $x\to x+1$, it is easily to get $f(x)=ax$ and putting back to the original, $a=1$. There are two solutions: $f(x)=0,f(x)=x$. $\square$