Is identity the only function $f$ on real line satisfying $ f(x +f(y) + yf(x)) = y +f(x) + xf(y) ,\forall x,y \in \mathbb R$? Let $f:\mathbb R \to \mathbb R$ be a function such that $ f(x +f(y) + yf(x)) = y +f(x) + xf(y) ,\forall x,y \in \mathbb R$ , then is it true that $f(x)=x,\forall x \in \mathbb R$ ? If not true in general , then what if we also assume $f$ is bijective , or say continuous ? 
Over $\mathbb C$ , $f(z)=\bar z$ is a bijective continuous , non-identity solution . Over $\mathbb R$ I can show that under the further assumption $\{f(x)/x : x\ne 0\}$ is countable, we must have $f$ is identity . With only the given functional equation , I can show that if $f(x_0)=0$ for some $x_0 $
 then $x_0=0$ , and then $f(f(x))=x,\forall x \in \mathbb R$ i.e. $f$ is bijective . But I can't figure out anything else 
Please help . Thanks in advance 
 A: If we assume continuity, then the identity function is the only possibility.
Edit: Here's a simpler argument based on Mohsen Shahriari's observation below that $f(-1)=-1$.  I'll leave my original argument after.
By Carl Schildkraut's argument in the comments, $f$ is an involution and so in particular a bijection.  So if we assume continuity, it must be either monotonically increasing or montonically decreasing.
Since $f(0)=0$ (as noted by Hagen von Eitzen above) and $f(-1)=-1$, it must be increasing.  But this together with the fact that it's an involution forces it to be the identity (otherwise, picking some $x$ with $f(x)\ne x$, it would have to decrease between $x$ and $f(x)$).
Original more complicated argument for why it can't be monotonically decreasing follows:
If $f$ were monotonically decreasing, there could be at most one fixed point (or it would have to increase between the two fixed points.)  We know zero is a fixed point.  But also, for any $x$, $x+f(x)+xf(x)$ is a fixed point, and so $x+f(x)+xf(x)=0$.  Solving, we get that, so long as $x\ne -1$, we have $f(x)=-\frac{x}{x+1}$.
Since $f$ is an involution on $\mathbb{R}$, whereas $x\mapsto -\frac{x}{x+1}$ is an involution on $\mathbb{R}\setminus\{-1\}$, we thus must have $f(-1)=-1$.
But this function isn't actually continuous, isn't actually monotonically decreasing, and doesn't satisfy the original functional equation.
