How can Chomsky hierarchy be applied to languages with alternated letters? I have the following grammar, which I know it is regular because it can be represented by a finite state automata:
\begin{array}{l}
\mathrm{S} \rightarrow \mathrm{X} \mid \mathrm{Y} \\
\mathrm{X} \rightarrow \mathrm{a} \mathrm{Y} \mathrm{b} \mid \mathrm{ab} \\
\mathrm{Y} \rightarrow \mathrm{b} \mathrm{X} \mathrm{a} \mid \mathrm{ba}
\end{array}
According to Chomsky hierarchy:

Type-3 grammars generate the regular languages. Such a grammar
restricts its rules to a single nonterminal on the left-hand side and
a right-hand side consisting of a single terminal, possibly followed
by a single nonterminal (right regular). Alternatively, the right-hand
side of the grammar can consist of a single terminal, possibly
preceded by a single nonterminal (left regular).

However, I do not find the hierarchy applied to the grammar. Am I missing something?
Every regular grammar produces a regular language.
But:
Not every grammar for a regular language is regular.
In other words, a regular language can be described by any of an infinitude of grammars, some of which are regular and some not.
In this case, you can easily create either a left- or a right-regular grammar for your language. For example:
\begin{array}{l}
\mathrm{S} \rightarrow \mathrm{X} \mid \mathrm{Y} \\
\mathrm{X} \rightarrow  \mathrm{X}  \mathrm{a}\mathrm{b} \mid \mathrm{ab} \\
\mathrm{Y} \rightarrow \mathrm{Y}   \mathrm{b}  \mathrm{a} \mid \mathrm{ba}
\end{array}
Of course, that transformation only preserves the language parsed, not the induced parse trees.