Example 6 Let $S$ be the set of all real numbers greater than -1, determine all functions $f: S \rightarrow S$, such that the following two conditions are satisfied:
(i) For all $x$ and $y$ in $S$, we have
$$
f(x+f(y)+x f(y))=y+f(x)+y f(x) \text {; }
$$
(ii) In each of the intervals $(-1,0)$ and $(0,+\infty)$, $\frac{f(x)}{x}$ is strictly increasing.
(35th IMO)
Solution: In condition (i), let $x=y$, we get
$$
\begin{array}{l}
f(x+f(x)+x f(x)) \\
=x+f(x)+x f(x) .
\end{array}
$$

Then $x+f(x)+x f(x)$ is a fixed point of $f$ (i.e., a $z$ such that $f(z)=z$).

Let $A=x+f(x)+x f(x)$, in equation (1), let $x=A$, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
f(A)=A, \\
f(A+f(A)+A f(A))=A+f(A)+A f(A)
\end{array}\right. \\
\Rightarrow f\left(A^{2}+2 A\right)=A^{2}+2 A .
\end{array}
$$

So, $A^{2}+2 A$ is also a fixed point of $f$.
If $A \in(-1,0)$, then
$$
A^{2}+2 A=(A+1)^{2}-1 \in(-1,0) \text {, }
$$

and $A^{2}+2 A \neq A$.
Thus, there are two fixed points in $(-1,0)$. This contradicts the fact that $\frac{f(x)}{x}$ is strictly increasing in $(-1,0)$.

If $A \in(0,+\infty)$, then $A^{2}+2 A \in(0,+\infty)$, $A^{2}+2 A \neq A$, which also contradicts the fact that $\frac{f(x)}{x}$ is strictly increasing in $(0,+\infty)$.
Therefore, $A=0$, i.e.,
$$
x+f(x)+x f(x)=0 \text {. }
$$

Solving for $f(x)=-\frac{x}{1+x}$.
Now, we verify that $f(x)=-\frac{x}{1+x}$ indeed satisfies the condition. Clearly, $\frac{f(x)}{x}=-\frac{1}{1+x}$ is strictly increasing in $S$.
For any $x, y \in S$, we have
$$
\begin{array}{l}
f(x+f(y)+x f(y)) \\
=f\left(x+\frac{-y}{1+y}+\frac{-y}{1+y} \cdot x\right) \\
=f\left(\frac{x-y}{1+y}\right)=-\frac{\frac{x-y}{1+y}}{1+\frac{x-y}{1+y}}=\frac{y-x}{1+x} . \\
y+f(x)+y f(x) \\
=y+\frac{-x}{1+x}+\frac{-x}{1+x} \cdot y=\frac{y-x}{1+x} .
\end{array}
$$

Thus, condition (i) holds.
Therefore, the function we seek is
$$
f(x)=-\frac{x}{1+x} .
$$
(To be continued)