The problem says that:
Let $\mathbb{R}$ be the set of real numbers. Find all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying the condition: $$f\big(xf(y)-y\big)+f(xy-x)+f(x+y)=2xy$$ for all $x,y\in \mathbb{R}$.
I am a little bit lost in the solution, I started to search some special values.
If $(x,y)=(0,0)$: $$\begin{align} f\big(0f(0)-0\big)+f(0)+f(0)=0 \\ 3f(0)=0 \end{align}$$ Thus: $f(0)=0$.
Is it right to continue and find some values by substitution or there is another technique?
Because I found that if $y=0$: $$f\big(xf(0)\big)+f(-x)+f(x)=0$$ therefore: $$\fbox{$f(x)=-f(-x)$}\tag1\label1$$ If $y=1$: $$f\big(xf(1)-1\big)+f(0)+f(x)=2$$ Giving $0$ to $x$: $$f(-1)=2$$ Using \eqref{1} we have: $$\begin{align} f(-1)&=-f(1) \\ f(1)&=-2 \end{align}$$ With all these values I conclude that: $$f : x \mapsto -2x$$ I feel that I'm wrong and I need another technique to solve this. But the condition is satisfied: Since $f(x)=-2x$ thus: $$\begin{align} f\big(xf(y)-y\big)&=-2\big(x\cdot (-2y)-y\big) \\ &=-2y+4xy\end{align}$$ and: $$f(xy-x)=-2xy+2x$$ Also: $$f(x+y)=-2x-2y$$ Thus: $$\begin{align}f\big(xf(y)-y\big)+f(xy-x)+f(x+y)&=2y+4xy-2xy+2x-2x-2y \\ &=2xy \end{align}$$ Nevertheless, I feel that there's another function, and this technique is wrong. Please help me, and thanks in advance.
let $P(x,y)$ be the assertion $$f\big(xf(y)-y\big)+f(xy-x)+f(x+y)=2xy\text.$$
$P(0,0)$ gives $f(0)=0$.
$P(x, 0)$: $f(-y)+f(y)=0$ therefore $f$ is odd.
$P(-x, -y)$ gives $$f\big(xf(y)+y\big)+f(xy+x)-f(x+y)=2xy\text.\tag{*}\label{*}$$
By plugging $x=1$ in \eqref{*}, we get $f\big(f(y)+y\big)=2y$, hence, $f$ is surjective.
$\therefore \exists a: f(a)=-1$.
$P(x,a)$: $f(-x-a)+f\big(x(a-1)\big)+f(x+a)=f\big(x(a-1)\big)=2ax$ (note that $f$ is odd).
$\therefore f\big(x(a-1)\big)=2ax$.
if $a=1$, $f(0)=2x$ which does not make sense. so $a\ne 1$, which implies that $f$ is linear. Putting back to the given assertion, we get $f(x)=x$.