= Solution
For $x$, there is an accepting <graph path>
$$
q_0\xrightarrow{1}q_1\xrightarrow{1}q_2
\xrightarrow{0}q_0\xrightarrow{0}q_0
\xrightarrow{1}q_1\xrightarrow{1}q_2
\xrightarrow{0}q_0\xrightarrow{\varepsilon}q_2.
$$
Its nonempty labels spell $1100110$, and its last <automaton state> is accepting. \b[Therefore $x$ is accepted.] The last <epsilon transition> matters: finishing the consumed input at $q_0$ does not prevent a subsequent move to $q_2$.
For $y$, track all possible <automaton states>, including their <epsilon closure>, after each input prefix:
|| Input prefix
|| Possible <automaton states> after <epsilon closure>
| $\varepsilon$
| $\{q_0,q_2\}$
| $1$
| $\{q_1\}$
| $11$
| $\{q_2\}$
| $110$
| $\{q_0,q_2\}$
| $1100$
| $\{q_0,q_2\}$
| $11001$
| $\{q_1\}$
The final <set> contains no accepting <automaton state>, and $q_1$ has no outgoing <epsilon transition>. \b[Therefore $y$ is rejected.]
$$
\boxed{x\in L(M),\qquad y\notin L(M).}
$$
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