Saudi Arabian Mathematical Competitions

Let denote by $P(x, y)$ the equation
$$f(x f(y)-y)+f(x y-x)+f(x+y)=2 x y.$$
$P(0, y)$ gives us $f(-y)+f(y)=0, \forall y$. Thus $f$ is an odd function.

$P(-1, y)$ follows
$$f(-f(y)-y)+f(-y+1)+f(-1+y)=-2 y.$$
From this, since $f$ is odd, we have $f(-f(y)-y)=-2 y$ and thus $f(f(y)+ y)=2 y$. So $f$ is surjective.

Since $f$ is surjective, there is real number $a$ such that $f(a)=-1$.

$P(x, a)$ gives us
$$f(-x-a)+f(a x-x)+f(x+a)=2 a x.$$
From this, again since $f$ is odd we have $f(a x-x)=2 a x$ for all $x \in \mathbb{R}$.\ (*)$

If $a=1$ then from above equation, we have $f(0)=2 x, \forall x \in \mathbb{R}$, which is a contradiction. Thus, $a \neq 1$.

Replace $x=\frac{t}{a-1}$ in $(*)$, we have
$$f(t)=\frac{2 a}{a-1} t=c t \text{ with } c=\frac{2 a}{a-1}.$$
Replace back into the original equation, we have
$$c(c x y-y)+c(x y-x)+c(x+y)=2 x y.$$
From which $\left(c^{2}+c-2\right) x y=0, \forall x, y \in \mathbb{R}$ which implies that $c \in\{1,-2\}$.

So $f(x)=x$ or $f(x)=-2 x$ for all $x \in \mathbb{R}$.

It is easy to check that these two functions satisfy the condition.