Example 1 Let $S$ denote the set of all real numbers greater than -1. Determine all functions: $S \rightarrow S$, satisfying the following two conditions:
(i) For all $x$ and $y$ in $S$,
$$
f(x+f(y)+x f(y))=y+f(x)+y f(x) ;
$$
(ii) In every interval within $-1 < x < 0$ and $0 < x$, $\frac{f(x)}{x}$ is strictly increasing.
From (i) we get
$$
f(x+f(x)+x f(x))=x+f(x)+x f(x) .
$$

For a fixed $x$, let $x+f(x)+x f(x)=c$, then the above equation becomes
$$
f(c)=c .
$$

Substituting $c$ into (1) and using (2), we get
$$
f\left(2 c+c^{2}\right)=2 c+c^{2} .
$$

Since $2+c>2+(-1)=1$, $2 c+c^{2}=c(2+c)$ has the same sign as $c$.
If $c>0$, then $2 c+c^{2}>c$, but (2) and (3) yield
$$
\frac{f\left(2 c+c^{2}\right)}{2 c+c^{2}}=\frac{f(c)}{c}=1,
$$

which contradicts the fact that $\frac{f(x)}{x}$ is strictly increasing for $x>0$.
If $c<0$, a similar contradiction arises.
Therefore, $c=0$. Thus, for all $x \in S$,
$$
x+f(x)+x f(x)=0 .
$$

That is,
$$
f(x)=-\frac{x}{x+1} .
$$

It is easy to verify that this function satisfies the requirements.
This solution cleverly uses the strict increase of $\frac{f(x)}{x}$ to quickly achieve the goal.