<s>[INST] 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
 [/INST] 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 $$g(ug(v))=vg(u).$$ 
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