Brazilian Math Olympiad

First, $f$ is bijective because, plugging $x=1$, we get $f(f(y)+f(1))=2f(1)+x$; $x+2f(1)$ spans all real numbers (so $f$ is surjective) and $f(x)=f(y)⟺f(x)+f(1)=f(y)+f(1)⟺f(f(x)+f(1))=f(f(y)+f(1))⟺2f(1)+x=2f(1)+y⟺x=y$ (so $f$ is injective). Choose $x\ne 0$ and $y$ such that $2f(x)+xy=f(x)⟺y=-\frac{f(x)}{x}$. By injectivity, $$xf\left(-\frac{f(x)}{x}\right)+f(x)=x⟺f\left(-\frac{f(x)}{x}\right)+\frac{f(x)}{x}=1(\ast )$$ Now, set $x=y=0$: $f(f(0))=2f(0)$. Let $a=f(0)$, so that $f(a)=2a$. If $a=0$, setting $y=0$ we get $f(f(x))=2f(x)$ and by surjectivity $f(x)=2x$ for all $x$. This function doesn't work (by testing), so $a$ can't be 0. So $a\ne 0$ and plugging $x=a$ in $(\ast )$, we obtain (recall that $f(a)=2a$): $$f(-2)+2=1⟺f(-2)=-1$$ Now set $y=0$: $f(f(x)-x)=2(f(x)-x)$. Now we find all real numbers $k$ such that $f(k)=2k$. If there is only one possible value for $k$, we are almost done because $f(x)-x=k⟺f(x)=x+k$ and by testing the function, $f(x)=x+1$. Let $m$ be such that $f(m)=k$ (it exists because $f$ is surjective). Plug $x=m$ and $y=0$: recalling that $f(0)=a$, $$f(mf(0)+f(m))=2f(m)⟺f(ma+k)=2k=f(k)$$ By injectivity, $$ma+k=k⟺ma=0⟺m=0$$ because we already proved that $a\ne 0$. So there is only one value for $k$, namely, $k=f(0)=a$. Then, we're done: the only function is $f(x)=x+1$.