For $x=y=0$ we obtain that $f(f(0))=0$ . For $x=0$ and $y=f(0)$ we obtain that $f(f(f(0)))=f(0)^2+f(0)+f(f(0))$ , so $f(0)^2=0$ and $f(0)=0$. For $x=0$ we obtain $f(f(y))=f(y)$ ,but $f$ is surjective , so we are done.

Plugging in $y = -1$ gives us $f(x + f(x - 1)) = f(-1) + 2x$ for all $x \in \mathbb{R}$. In particular, $f$ is surjective. Plugging in $x = 0$, we get \[ f(f(y)) = (y + 1)f(0) + f(y) \tag{1} \]for all $y \in \mathbb{R}$. Since $f$ is surjective, there exists $t \in \mathbb{R}$ such that $f(t) = 0$. Then, plugging $y = t$ into (1), we find $tf(0) = 0$, so either $f(0) = 0$ or $t = 0$. Either way, $f(0) = 0$. Finally, (1) becomes $f(f(y)) = f(y)$ for all $y$ real. Since $f$ is surjective, we get $\boxed{f(x) = x \; \forall x \in \mathbb{R}}$.

parola wrote: For $x=y=0$ we obtain that $f(f(0))=0$ . For $x=0$ and $y=f(0)$ we obtain that $f(f(f(0)))=f(0)^2+f(0)+f(f(0))$ , so $f(0)^2=0$ and $f(0)=0$. For $x=0$ we obtain $f(f(y))=f(y)$ ,but $f$ is surjective , so we are done. From $x=y=0$ we obtain $f(f(0)) = 2f(0),$ which is not what you wrote. Also, you haven't proven that $f$ is surjective.

It is evident, that is surjective. It is enought to fix the y.

$P\left(\frac{x-f(-1)}2,-1\right)\Rightarrow f\left(\frac{x-f(-1)}2+f\left(\frac{x-f(-1)}2-1\right)\right)=x\Rightarrow f$ is surjective $P\left(0,\frac{-f(-1)}2+f\left(\frac{-f(-1)}2-1\right)\right)\Rightarrow\left(\frac{-f(-1)}2+f\left(\frac{-f(-1)}2-1\right)\right)f(0)=0\Rightarrow f(0)=0$ $P(0,x)\Rightarrow f(f(x))=f(x)\Rightarrow\boxed{f(x)=x}$ by surjectivity, works.

Here goes my long-winded solution... The only function is $f\equiv x$. It is easy to check that this works. Now we prove that it is the only one. As per usual, denote the assertion by $P(x,y)$. First, note that $P(x,-1)$ gives that $f$ is surjective. Therefore, there exists an $a$ such that $f(a)=0$. Claim: $f(0)=0$. $P(a,0)$ gives us that $f(a+f(a))=f(a)+f(0)+a\implies f(a)=f(0)+a\implies f(0)=-a$. Then $P(0,a)$ gives that $f(f(a))=af(0)+f(0)\implies 0=(a+1)f(0)$. Therefore, either $a=-1\implies f(0)=1$ or $f(0)=0$. We show the former is impossible. Suppose $f(0)=1$. Then $P(0,y)$ gives us that $f(f(y))=yf(0)+f(0)+f(y),$ so that because $f(0)=1,$ then $f(f(y))=f(y)+y+1$. Therefore, if $f(x_1)=f(x_2),$ then $f(f(x_1))=f(x_1)+x_1+1=f(f(x_2))=f(x_2)+x_2+1\implies x_1=x_2,$ so $f$ is injective. Using the classic symmetry trick in the original equation, we get that $f(x+f(x+y))+xy=yf(x)+f(x)+f(y)+x$ and $f(y+f(x+y))+xy=xf(y)+f(y)+f(x)+y,$ and subtracting the two gives $f(x+f(x+y))-f(y+f(x+y))=yf(x)-xf(y)+x-y$. Setting $x=0$ in this equation gives us that $f(f(y))=f(y+f(y))$. Because $f$ is injective we get that $f(y)=y+f(y),$ which is clearly not true for all $y$(pick any nonzero $y$). Contradiction, so $f(0)=0$. From here it is easy. In the equation we got, namely, $f(x+f(x+y))-f(y+f(x+y))=yf(x)-xf(y)+x-y,$ we can substitute $y=-x$. This simplifies to $(x+1)f(x)=f(-x)(-x+1)+2x$. Now all we have to do is find $f(-x)$ in terms of $f(x),$ and then we'll be done. Fortunately, this is just putting in $y=-x$ in the original equation, and from there we can easily solve. Do note that we must divide by $-x+1$ in the process, so we still have to confirm that $f(x)=x$ for $x=1,$ but this is easy. We have that $2f(1)=2\implies f(1)=1$. We are done. Note: Somehow I forgot that subjectivity existed when I got $f(f(y))=f(y)$!!