Example 11 Let $x>0, y>0, \sqrt{x}(\sqrt{x}+2 \sqrt{y})$ $=\sqrt{y}(6 \sqrt{x}+5 \sqrt{y})$. Find the value of $\frac{x+\sqrt{x y}-y}{2 x+\sqrt{x y}+3 y}$.
Solution: From the given, we have
$$
(\sqrt{x})^{2}-4 \sqrt{x} \sqrt{y}-5(\sqrt{y})^{2}=0,
$$

which is $(\sqrt{x}-5 \sqrt{y})(\sqrt{x}+\sqrt{y})=0$.
$$
\begin{array}{l}
\because \sqrt{x}+\sqrt{y}>0, \\
\therefore \sqrt{x}-5 \sqrt{y}=0,
\end{array}
$$

which means $x=25 y$.
Substituting $x=25 y$ into the original fraction, we get
the original expression $=\frac{1}{2}$.