TITLE: Find the function equation $f(x+f(y)+yf(x))=y+f(x)+xf(y)$
QUESTION [6 upvotes]: find  all functions $f:\Bbb R\to \Bbb R$ and such for any $x,y\in\Bbb R$,and such
$$f(x+f(y)+yf(x))=y+f(x)+xf(y)$$

I have proven that $$f(f(x))=x$$

proof:Let $y=0$,we have
$$f(x+f(0))=f(x)+xf(0)$$
let $x=0,y\to x$,we have
$$f(f(x)+xf(0))=x+f(0)$$
so we have
$$f(f(x+f(0))=x+f(0)$$
then $$f(f(x))=x$$
and $f(x)=x$ is one solution but I can't argue this to be the only solution.
It doesn't seem easy.Thank for you help!

REPLY [2 votes]: There is a proof given by Professor Wu Wei-chao in 2001, you may download the paper here. 
Here is the overview of the proof:

$f$ is one-to-one, and $$f(0)=0,\tag{2}$$
$\forall y\in\Bbb R\setminus \{0\}$, $$f\left(-\frac y{f(y)}\right)=-\frac{f(y)}y,\qquad f\left(-\frac{f(y)}y\right)=-\frac y{f(y)},\tag{8,9}$$ 
$$f(\pm1)=\pm1,\tag{11,12}$$
$\forall x\in\Bbb R$, 
$$f(x+f(x)+1)=f(x)+x+1,\qquad f(x-f(x)-1)=-x+f(x)-1\tag{13,14}$$
$$f(\pm2)=\pm2,\tag{23,29}$$
$$(f(\pm\sqrt2))^2=2,\qquad(f(\pm\frac 1{\sqrt2}))^2=\frac12,\tag{42,46}$$
$\forall x\in\Bbb R$, 
$$f(3x-4)=-3f(-x)-4.\tag{47}$$
Introduce 5 lemmas,

Lemma 4. $\forall a\in \Bbb R$, if $$\tag{58}f(-x-a)=-f(x)-a,\qquad (\forall x\in\Bbb R)$$ and $$\tag{59}f(x-f(x)-\dfrac {a^2}4)=f(x)-x-\dfrac {a^2}4,\qquad (\forall x\in\Bbb R)$$ 
  then 
  $$f(x)=x\qquad (\forall x\in\Bbb R).$$

Saparate into 3 cases and finish the proof.


Case 1: $f(\pm\sqrt 2)=\pm\sqrt 2\implies f(x)=x$ ($\forall x\in\Bbb R$).
Case 2: $f(\pm\sqrt 2)=\mp\sqrt 2, f(\pm\frac 1{\sqrt2})=\pm\frac 1{\sqrt2}\implies$ contradition to injectivitiy of $f$.
Case 3: $f(\pm\sqrt 2)=\mp\sqrt 2, f(\pm\frac 1{\sqrt2})=\mp\frac 1{\sqrt2}\implies$ contradition to $f(\sqrt 2)=-\sqrt 2$.


Wu also generalized the result.

Generalization: Suppose $F$ is a field or a ring, $\Bbb Z\subseteq F\subseteq \Bbb C$, and $f:F\to F$ satisifies
  $$f(x+f(y)+yf(x))=y+f(x)+xf(y),$$

if $F=\Bbb R$, then $f(x)=x$ ($\forall x\in\Bbb R$);
if $F=\Bbb C$, then $f(x)=x$ ($\forall x\in\Bbb C$) or $f(x)=\overline x$ ($\forall x\in\Bbb C$);
if $F=\Bbb Q$, then $f(x)=x$ ($\forall x\in\Bbb Q$);
if $F=\Bbb Z$, then $f(x)=x$ ($\forall x\in\Bbb Z$);
if $F=\{a+br\mid \forall a,b\in\Bbb Q\}$, where $r$ is fixed irrational and $r^2\in\Bbb Q$, then $f(x)=x$ ($\forall x\in F$) or $f(x)=\overline x$ ($\forall x\in F$), here we define $\overline{a+br}=a-br$;
if $F=\{a+br\mid \forall a,b\in\Bbb Z\}$, where $r$ is fixed irrational and $r^2\in\Bbb Z$, then $f(x)=x$ ($\forall x\in F$) or $f(x)=\overline x$ ($\forall x\in F$), here we define $\overline{a+br}=a-br$.