TITLE: functional equation of type $f(x+f(y)+xf(y)) = y+f(x)+yf(x)$
QUESTION [5 upvotes]: If $f:\mathbb{R}-\{-1\}\rightarrow \mathbb{R}$ and $f$ is a differentiable function that satisfies $$f(x+f(y)+xf(y)) = y+f(x)+yf(x)\forall x,y \in \mathbb{R}-\{-1\}\;,$$ Then value of $\displaystyle 2016(1+f(2015)) = $

$\bf{My\; Try::}$ Using partial Differentiation, Differentiate w r to $x$ and $y$ as a constant
$$f'(x+f(y)+xf(y)) \cdot (1+f(y)) = f'(x)+yf'(x)$$
Similarly Differentiate w r to $y$ and $x$ as a constant
$$f'(x+f(y)+xf(y))\cdot (f'(y)+xf'(y)) = 1+f(x)$$
Now Divide these two equation, We get $$\frac{1+f(y)}{(1+x)f'(y)} = \frac{(1+y)f'(x)}{1+f(x)}$$
Now How can i solve it after that, Help required, Thanks

REPLY [2 votes]: Let $x=0$ then 
$$f(f(y))=y\,\big(\,f(0)+1\big)+f(0)\tag{1}$$
and from this we get $f(f(0))=f(0)$ and $f(f(1))=2f(0)+1$. Let $y=0$ and $x=1$ in original equation then 
$$f\big(1+2f(0)\big)=f(1)\tag{2}$$
applying $\,f$ we get 
$$f(f(1+2f(0)))=f(f(1))=2f(0)+1$$
Substituting $y=1+2f(0)$ in $(1)$ we obtain
$$f(f(1+2f(0)))=(1+2f(0))(f(0)+1)+f(0)=1+3f(0)+2(f(0))^2,$$
and from the last two equations we get
$$f(0)\big(1+f(0)\big)=0.$$
If $f(0)=-1$, then $f(f(0))=f(-1)$ which is not defined but we know that $f(f(0))=f(0)$. Thus $f(0)=0$ and from $(1)$ $\:f(f(x))=x$. For the next step see This post