functional equation of type $f(x+f(y)+xf(y)) = y+f(x)+yf(x)$ 
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
 A: Your last equation tells us that
$$
\frac{(1+x)f'(x)}{1+f(x)}
$$
has to be constant, and that this constant has to be equal to its reciprocal, i.e., it is $\pm1$.
Integration gives
$$
\ln|1+f(x)|=\pm \ln|1+x|+C
$$
and thus
$$
f(x)=C(1+x)-1\text{ or } f(x)=\frac{C}{1+x}-1
$$
This now needs to be tested against the original functional equation.

The equation can be rewritten as $1+f((1+x)(1+f(y))-1)=(1+y)(1+f(x))$. 
\begin{array}{rl|l}
f(x)&=C(1+x)-1&=\dfrac{C}{1+x}-1\\ \hline
(1+x)(1+f(y))&=C(1+x)(1+y) & =C\dfrac{1+x}{1+y}\\
1+f((1+x)(1+f(y))-1)&=C^2(1+x)(1+y) & =\dfrac{1+y}{1+x}\\
(1+y)(1+f(x))&=C(1+y)(1+x) & = C\dfrac{1+y}{1+x}
\end{array}
which gives $f(x)=-1$, $f(x)=x$ and $f(x)=\dfrac{1}{1+x}-1=-\dfrac{x}{1+x}$ as valid solutions.
Thus the possible values for $(1+x)(1+f(x))$, for $x=2015$ or elsewhere, are $0$, $(1+x)^2$ or $-1$.
PS: As discussed with Patrick Stevens and egreg, for the function $f\equiv -1$ the functional equation is nowhere defined, as the argument of $f$ on the left would be $x(1+f(y))+f(y)=-1$ and thus outside the domain of $f$.

If one wants to simplify the functional equation first, set $g(x)=1+f(x-1)$, $u=1+x$, $v=1+x$ then the reduced form is $$g(ug(v))=vg(u).$$ 
A: Summing up the comments; both the identity map and the constant map $-1$ are differentiable functions that satisfy the functional equation. Hence the value of $2016(1+f(2015))$ is not uniquely determined by the given data.
A: (Work in progress; I was unexpectedly called away. The aim of this post is to do it without assuming differentiability.)
Roots
To find the roots of $f$: suppose $f(y) = 0$.
Then $f(x) = y+f(x)+y f(x)$ and so $(1+f(x))y = 0$; therefore $f(x) = -1$ for all $x$, or $y=0$.
So we have two possibilities: $f(x) = -1$ for all $x \not = -1$, or $f(0) = 0$.
But if $f(x) = -1$ for all $x \not = -1$, then letting $x=0$ we obtain $$f(-1) = -1$$
which is a contradiction to the undefinedness of $f(-1)$.
(There's a slightly simpler contradiction if we simply suppose there is any $y$ with $f(y) = -1$, as egreg points out.)
Therefore $f$ has exactly one root, and it is at $0$.
(Strictly, we should go back through this, and verify that we never tried to give $f$ the argument $-1$. The easiest way to do this is to let $x$ be some real such that $f(x) \not = -1$, and then just go through the same proof again.)
Using the root
Letting $x=0$, we see that $f(f(y)) = y$ for all $y$.
Letting $y=f(x)$, we obtain $$f(2x+x^2) = 2f(x)+f(x)^2$$
so, if $x=-2$, we get $f(0) = 2 f(-2) + f(-2)^2$; that is, $f(-2) = 0$ or $f(-2) = -2$.
We already know $f$ has exactly one root, so $f(-2) = -2$.
Letting $x = f(y)$, we obtain $$f(2f(y)+f(y)^2) = 2y+y^2$$
so if $f(y) = -2$, we obtain $y = 0$ (contradiction) or $y=-2$.
Therefore $f$ hits $-2$ at exactly one input: namely $-2$.
Letting $y=-2$, then, we obtain $f(-x-2) = -2+f(x)-2f(x)$, so $$f(-x-2) = -f(x)-2$$
and so the behaviour of the function is determined precisely by its behaviour on $x>-1$.
Actually, under the assumption that there is $x$ such that $f(x) = 1$, we obtain $f(x + f(y) + x f(y)) = 1 + 2 y$ and hence (by applying $f$ to both sides) $$x + f(y) + x f(y) = f(1 + 2 y)$$
whence (substituting $z=1+2y$) we get $$f(z) = f\left(\frac{z-1}{2}\right) + x f\left(\frac{z-1}{2}\right) + x$$
so any open interval $(-1, r)$ determines the behaviour of $f$ completely.
TODO: complete this. As LutzL points out, there is more than one function satisfying the recurrence :(
A: 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
