Example 1 Given $a^{2}+b^{2}=6 a b$, and $a>b>0$. Then $\frac{a+b}{a-b}=$ $\qquad$
$(2001$, Beijing Middle School Mathematics Competition (Grade 8))
Let $a=x+y, b=x-y$. Substituting into the given equation and simplifying, we get $4 x^{2}=8 y^{2}$.
Since $a>b>0$, it follows that $x>y>0$.
Thus, $\frac{x^{2}}{y^{2}}=2 \Rightarrow \frac{x}{y}=\sqrt{2}$.
Notice that $a+b=2 x, a-b=2 y$.
Therefore, $\frac{a+b}{a-b}=\frac{2 x}{2 y}=\frac{x}{y}=\sqrt{2}$.