Notice
$$\begin{array}{c} y=\frac{\frac{1}{b}}{\left(\frac{a}{x_{1}}+1\right)\left(\frac{x_{1}}{x_{2}}+1\right) \cdots\left(\frac{x_{n-1}}{x_{n}}+1\right)\left(\frac{x_{n}}{b}+1\right)} . \\ \text { Let } a_{1}=\frac{a}{x_{1}}, a_{2}=\frac{x_{1}}{x_{2}}, \cdots \cdots a_{n}=\frac{x_{n-1}}{x_{n}}, \end{array}$$
$a_{n+1}=\frac{x_{n}}{b}$. Then $a_{1} a_{2} \cdots a_{n+1}=\frac{a}{b}$.
By the extended version 1, we know
$$\begin{array}{l} \left(\frac{a}{x_{1}}+1\right)\left(\frac{x_{1}}{x_{2}}+1\right) \cdots\left(\frac{x_{n-1}}{x_{n}}+1\right)\left(\frac{x_{n}}{b}+1\right) \\ =\prod_{i=1}^{n+1}\left(a_{i}+1\right) \geqslant\left(\sqrt[n+1]{\frac{a}{b}}+1\right)^{n+1}, \end{array}$$

Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{n+1}=\sqrt[n+1]{\frac{a}{b}}$, that is,
$$\frac{a}{x_{1}}=\frac{x_{1}}{x_{2}}=\cdots=\frac{x_{n-1}}{x_{n}}=\frac{x_{n}}{b}=\sqrt[n+1]{\frac{a}{b}},$$

which means $x_{k}=a\left(\frac{b}{a}\right)^{\frac{k}{n+1}}$ for the equality to hold.
Thus, $y \leqslant \frac{\frac{1}{b}}{\left(\sqrt[n+1]{\frac{a}{b}}+1\right)^{n+1}}=\frac{1}{(\sqrt[n+1]{a}+\sqrt[n+1]{b})^{n+1}}$.
Equality holds if and only if $x_{k}=a\left(\frac{b}{a}\right)^{\frac{k}{n+1}}(k=1,2, \cdots, n)$

when $y$ reaches its maximum value $\frac{1}{(\sqrt[n+1]{a}+\sqrt[n+1]{b})^{n+1}}$.