Example 13 Let $f(x)$ be an increasing function defined on $(0,+\infty)$ and satisfying $f\left(\frac{x}{y}\right)=f(x)-f(y), f(3)=1$. Try to solve the functional inequality
$$
f(x)-f\left(\frac{1}{x-5}\right) \geqslant 2 .
$$
Given that $f(x)-f\left(\frac{1}{x-5}\right) \geqslant 2$, we have $f\left[\frac{x}{(x-5)^{-1}}\right] \geqslant 2 f(3)$.
That is, $f\left[\frac{x(x-5)}{3}\right] \geqslant f(3)$.
Since $f(x)$ is an increasing function on $(0,+\infty)$, we have
$$
\left\{\begin{array}{l}
x>0 \\
x-5>0 \\
\frac{x(x-5)}{3} \geqslant 3
\end{array}\right.
$$

Solving this, we get $x \geqslant \frac{5+\sqrt{61}}{2}$.
Analysis If the inequality $f(x)-f\left(\frac{1}{x-5}\right) \geqslant 2$ in this example is generalized to $f(x)-f\left(\frac{p}{x-5}\right) \geqslant 2$, then the solution to this inequality is:
When $p>0$, $x>\frac{1}{2}(5+\sqrt{25+36 p})$;
When $-\frac{25}{36}<p<0$, the solution is
$$
\frac{1}{2}(5-\sqrt{25+36 p})<x<\frac{1}{2}(5+\sqrt{25+36 p}) .
$$