Example 4 (Adapted from the 26th International Mathematical Olympiad 1994) Let $f:(-1,+\infty) \rightarrow (-1,+\infty)$ be a continuous and monotonic function, with $f(0)=0$, and satisfying
$$
f[x+f(y)+x f(y)] \geqslant y+f(x)+y f(x) \text {, for } \forall x, y \in(-1,+\infty),
$$

Find $f(x)$.
Solve: Since $f(x)$ is a continuous monotonic function, and $f(0)=0$, $f(x)$ has opposite signs on either side of the zero point, and as $y \rightarrow 0$, $f(y) \rightarrow 0, x+f(y)+x f(y)-x \rightarrow 0$.
$$
f[x+f(y)+x f(y)]-f(x) \geqslant y[1+f(x)],
$$

When $y>0$,
$$
\frac{f[x+f(y)+x f(y)]-f(x)}{x+f(y)+x f(y)-x} \cdot \frac{f(y)+x f(y)}{y} \geqslant 1+f(x),
$$

Since $1+x>0$, we have $\frac{f[x+f(y)+x f(y)]-f(x)}{x+f(y)+x f(y)-x} \cdot \frac{f(y)}{y} \geqslant \frac{1+f(x)}{1+x}$,
Let $y \rightarrow 0$, we get $f^{\prime}(x) f^{\prime}(0) \geqslant \frac{1+f(x)}{1+x}$,
When $-1 < y < 0$, we get $\frac{f[x+f(y)+x f(y)]-f(x)}{x+f(y)+x f(y)-x} \cdot \frac{f(y)}{y} \leqslant \frac{1+f(x)}{1+x}$, let $y \rightarrow 0$, we get $f^{\prime}(x) f^{\prime}(0) \leqslant \frac{1+f(x)}{1+x}$,
Thus, we have $f^{\prime}(x) f^{\prime}(0)=\frac{1+f(x)}{1+x}$,
Let $f^{\prime}(0)=a$, the above equation becomes $a(1+x) f^{\prime}(x)=1+f(x)$,
When $f^{\prime}(0)=0$, $1+f(x)=0$, so $f(x) \equiv-1$, which is not a monotonic function, hence discard.
When $f^{\prime}(0) \neq 0$, the above equation becomes
$$
\begin{array}{l}
a(1+x) f^{\prime}(x)-f(x)=1, \\
\text { Thus, }\left[\left(\frac{1}{1+x}\right)^{\frac{1}{*}} f(x)\right]^{\prime}=\left[(1+x)^{-\frac{1}{*}} f(x)\right]^{\prime} \\
=(1+x)^{-\frac{1}{*}} f^{\prime}(x)-\frac{1}{a}(1+x)^{-\frac{1}{a}-1} f(x) \\
=\frac{1}{a}\left(\frac{1}{1+x}\right)^{1+\frac{1}{x}}\left[a(1+x) f^{\prime}(x)-f(x)\right] \text {, } \\
{\left[\left(\frac{1}{1+x}\right)^{\frac{1}{4}} f(x)\right]^{\prime}=\frac{1}{a}\left(\frac{1}{1+x}\right)^{1+\frac{1}{x}},} \\
\left(\frac{1}{1+x}\right)^{\frac{1}{4}} f(x)=\int \frac{1}{a}\left(\frac{1}{1+x}\right)^{1+\frac{1}{4}} d x=-\left(\frac{1}{1+x}\right)^{\frac{1}{x}}+c, \\
f(x)=c(1+x)^{\frac{1}{n}}-1 \text {. } \\
\end{array}
$$

By $f(0)=0$, we get $c=1$, by $f^{\prime}(0)=a$, substituting $x=0$ into (1), we get $a^{2}=1$, i.e., $a=-1$ or $a=1$. When $a=-1$, $f(x)=(1+x)^{-1}-1=-\frac{x}{1+x}$;
When $a=1$, $f(x)=(1+x)-1=x$.
In summary, the solutions to the given functional inequality are $f(x)=-\frac{x}{1+x}$; $f(x)=x$.
Verification: Both solutions satisfy the original functional inequality.