A generic approach to solve such questions would be to come up with the corresponding regular languages and compare the two.
Ardens theorem could be used to achieve the objective. Moreover, some conclusions could be drawn without even deriving the languages completely, solely from the intermediary equations.
Constructing equations for every state -: X = Z0 +X1, Y = X0 +Y0 +Z0 && Z = X0 + Y1 + Z1.
Clearly, regular languages L1 and L2 couldn’t be same. Now, to check which of the other three options are correct, we need to find : 1) a string which is accepted by L1 but not by L2. 2) a string which is accepted by L2 but not by L1. If there exists string 1) but not 2) then L2 ⊂ L1 and likewise. Trying our hands at the given DFA, 1) string 00 is accepted by L1 but not by L2 and 2) string 01 is accepted by L2 but not by L1. Hence, neither of the two languages our subset of the other.
L1 can have 00 string while L2 can't. L2 can have 01 while L1 can't None of them can be either equal or proper subset of each other.
So Ans. D.