Example 1 Given $x, y \in \mathbf{R}_{+}$, and $\frac{1}{x}-\frac{1}{y}=1$, find the maximum value of $x-y$.
From the second conclusion above, we get
$$\begin{aligned}
x-y & =(x-y)\left(\frac{9}{x}-\frac{1}{y}\right) \\
& =\left[(\sqrt{x})^{2}-(\sqrt{y})^{2}\right]\left[\left(\frac{3}{\sqrt{x}}\right)^{2}-\left(\frac{1}{\sqrt{y}}\right)^{2}\right] \\
& \leqslant(3-1)^{2}=4
\end{aligned}$$

That is, \( x-y \leqslant 4 \) (when and only when \(\sqrt{x} \cdot \frac{1}{\sqrt{y}}=\sqrt{y} \cdot \frac{3}{\sqrt{x}}, \frac{9}{x}-\frac{1}{y}=1\), i.e., \(x=6, y=2\), then \(x-y=4\)), so the maximum value of \(x-y\) is 4.