Example 11 (2005 China Southeast Mathematical Olympiad Summer Camp Test Question) Find the function $f: \mathbf{K} \rightarrow \mathbf{R}$, satisfying:
(1) $x(f(x+1)-f(x))=f(x), \forall x \in \mathbf{R}$ and
(2) $|f(x)-f(y)| \leqslant|x-y|, \forall x, y \in \mathbf{R}$.
Let $x=0$, from (1) we get $f(0)=0$. When $x \neq 0, -1$, (1) can be rewritten as $\frac{f(x+1)}{x+1}=\frac{f(x)}{x}$.

In particular, for $x>0$, and $n \in \mathbf{N}^{*}$, we have $\frac{f(x+n)}{x+n}=\frac{f(x)}{x}$,
then $|x-y| \geqslant|f(x+n)-f(y+n)|=\left|\frac{x+n}{x} f(x)-\frac{y+n}{y} f(y)\right|$
$$
=\left|f(x)-f(y)+n\left(\frac{f(x)}{x}-\frac{f(y)}{y}\right)\right| .
$$

Thus $\frac{|x-y|}{n} \geqslant\left|\frac{f(x)-f(y)}{n}+\left(\frac{f(x)}{x}-\frac{f(y)}{y}\right)\right|$.
Let $n \rightarrow+\infty$, we get $\frac{f(x)}{x}=\frac{f(y)}{y}, \forall x, y>0$,
i.e., $f(x)=k x, x>0$.