If L is regular language then each equivalence class is also regular?
Given a DFA $D$ and a word $x$, let $D(x)$ denote the state of $D$ after reading $x$.

Let $D$ be a minimal DFA for $L$ and $x,y \in \Sigma^*$.
If $D(x) \neq  D(y)$ then $x$ and $y$ have a distinguishing extension.
Indeed, if $x$ and $y$ had no distinguishing extension, it would be possible to construct a DFA $D'$ with one less state by merging the states $D(x)$ and $D(y)$ of $D$.

On the converse, if $D(x)=D(y)$ then $x$ and $y$ have no distinguishing extension (since the states reached by $D$ from $D(x)$ and $D(y)$ after reading any string $z$ must coincide).

Then, $x$ and $y$ are in the same equivalence class iff $D(x)=D(y)=q$, for some state $q$ of $D$.

Given a state $q$ of $D$, the language of all the words $x \in \Sigma^*$ such that $D(x)=q$ is regular. To see this, consider the DFA obtained from $D$ by changing the set of final states to $\{q\}$.