IMO 1994- Problem 5

Problem: Let $\mathbb{S}$ be the set of real numbers strictly greater than $-1$. Find all functions $f:\mathbb{S}\to \mathbb{S}$ satisfying the two conditions:
(i) $f(x)+f(y)+xf(y)=y+f(x)+yf(x)$ for all $x$ and $y$ in $\mathbb{S}$.
(ii) $\frac{f(x)}{x}$ is strictly increasing on each of the intervals $-1<x<0$  and $0<x$.
Solution: Condition (ii) implies that $f(x)=x$ has at most three solutions, one in $(-1,0),$ one equal to $0$ and the third in $(0,\infty)$.
Suppose $f(u)=u$ for some $u\in (-1,0)$. Setting $x=y=u$ in (i), we get $f(u^2+2u)=u^2+2u\in (-1,0)$.
This means $u^2+2u=u$. But then $u\notin (-1,0)$. The case $f(v)=v$ for some $v>0$ leads to a similar contradiction.
  However, $f(x+(1+x)f(x))=x+(1+x)f(x)$ for all $x\in \mathbb{S}$. So we have $x+(1+x)f(x)=0$ which gives $f(x)=-\frac{x}{1+x}$.
It's routine to check that $f(x)=-\frac{x}{1+x}$ satisfies the desired property.
Source:http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf