[Example 1.6.4] Find all functions $f, g, h: \mathbf{R} \rightarrow \mathbf{R}$ that satisfy the following condition:
$$
f(x)-g(y)=(x-y) h(x+y), \quad \forall x, y \in \mathbf{R} .
$$
Here involves three functions $f(x), g(x)$, and $h(x)$. First, we need to discover their relationships by substituting special values. Since equation (1) contains the factor $x-y$, the first idea is to let $x-y=0$.
In equation (1), let $x=y$, then
$$
f(x)-g(x)=0 .
$$

So
$$
f(x)=g(x) .
$$

Thus, the problem is reduced to finding two functions $f(x)$ and $h(x)$.
Equation (1) can be rewritten as
$$
f(x)-f(y)=(x-y) h(x+y) .
$$

In equation (2), we take the special value $y=0$, then equation (2) becomes
$$
f(x)=x h(x)+f(0) .
$$

Using equation (3), equation (2) can be transformed into (eliminating $f(0)$)
$$
x h(x)-y h(y)=(x-y) h(x+y) \text {. }
$$
Equation (4) now involves only one function $h(x)$, so we can focus on solving $h(x)$.

For equation (4), we perform the following special value substitution, i.e., replace $(x, y)$ in equation (4) with $(x+y, -y)$, which gives
$$
(x+y) h(x+y)+y h(-y)=(x+2 y) h(x),
$$

To find $h(x)$, we first eliminate $h(x+y)$ from equations (4) and (5). For this, we can multiply equation (4) by $(x+y)$ and equation (5) by $(x-y)$, yielding
$$
2 y^{2} h(x)=(x+y) y h(y)-(x-y) y h(-y) .
$$

For equation (6), we take another special value for $y$, let $y=1$, then equation (6) becomes
$$
h(x)=\frac{1}{2}[(x+1) h(1)-(x-1) h(-1)]
$$

That is
$$
\begin{aligned}
h(x) & =\frac{1}{2} x[h(1)-h(-1)]+\frac{1}{2}[h(1)+h(-1)] \\
& =a x+b .
\end{aligned}
$$

where $a=\frac{1}{2}[h(1)-h(-1)], b=\frac{1}{2}[h(1)+h(-1)]$.
Substituting (7) into (3) gives
$$
\begin{aligned}
f(x) & =x(a x+b)+f(0) \\
& =a x^{2}+b x+c .
\end{aligned}
$$

where $c=f(0)$.
Thus
$$
\begin{array}{c}
f(x)=g(x)=a x^{2}+b x+c, \\
h(x)=a x+b .
\end{array}
$$

From the derivation process, it can be seen that for any real triplet $(a, b, c)$, the functions $f(x), g(x), h(x)$ defined by (8) and (9) all satisfy the given condition (1). Therefore, the set of all such $f(x), g(x), h(x)$ is the solution.