How can one show that $f(0)=0$ for $f$ satisfying certain conditions? Given the functional equation 
$$f(x+(1+x)f(y))=y+(1+y)f(x)$$
Such that $f:(-1,\infty) \to (-1,\infty)$ and the function $g(x):=\frac{f(x)}{x}$ is strictly increasing in $I=(-1,0)\cup(0,+\infty),$ how can one show that for every $t \in I$ we have 
$f(t)\neq t$?
Plugging $x=y=0,$ that makes $f(f(0))=f(0).$
Does that mean that $f(0)=0$?
Making $x=y,$ $f(x+(1+x)f(x))=x+(1+x)f(x)$ so we make $x+(1+x)f(x)=t$ that makes $f(t)=t$ but we can't make $t=0$ because $t$ is not in $I$? So do I have to use the fact that $\frac{f(x)}{x}$ is increasing?
Then after all this work we have to deduce that $f(0)=0$ and $f(x)=\frac{-x}{x+1}$ for $x \in (-1,\infty).$
 A: I think that this will do the trick for showing that f(0) = 0:
Let's start by setting $x = 0$. We get 
$$f(f(y)) = f(0) + (1+f(0))y.$$
By setting $y = 0$, we get 
$$f(x) = f(x + (1+x)f(0)).$$
Apply now $f$ to expression above and use the first equation:
$$f(0) + (1+ f(0))x = f(0) + (1+f(0))(x + (1+x)f(0)).$$
This simplifies to 
$$(1+x)(1+f(0))f(0) = 0$$
and thus $f(0) = 0$. We can also conclude that $f(f(y)) = y $, so f is its' own inverse. 

Let suppose that there is numbers $a,b$ such that $a < b$ and $f(a) = b$ and $f(b) = a$ ($f$ is its' own inverse). Since function
$$ g(x) = \frac{f(x)}{x}$$
is increasing by the hypothesis.  This means that $g(a) \le g(b)$, so 
$$\frac{f(a)}{a} \le \frac{f(b)}{b} \iff  \frac{b}{a} \le \frac{a}{b} \iff$$
$$\frac{b^2 - a^2}{ab} \le 0 \iff \frac{(b-a)(b+a)}{ab} \le 0 \iff $$
$$ \frac{(b+a)}{ab} \le 0. $$
so we can say that there can't be such $a,b$ with same sign. So $x$ and $f(x)$ must have  different signs. 
As OP noticed the equation 
$$ f(x+(1+x)f(x)) = x + (1+x)f(x) $$
holds. Let denote $x+ (1+x)f(x)$ as $t$. Then the equation says $f(t) = t$ but $f(t)$ and $t$ have different signs, therefore, $t$ must be zero. Thus, we conclude that $x + (1+x)f(x) = 0$ and the function $f$ is:
$$ f(x) = -\frac{x}{1+x}.$$
