Why is this language involving reversal regular?
Notice that big unconstrained $z$ in the middle. You don't need the DFA to check anything if there is a way to decompose the word that dumps almost everything into $z$.

In particular, any word of the form $a b z b a$ where $a$ and $b$ are letters and $z$ is an arbitrary non-empty word is in $L$ (with words of length 1 for $x$ and $y$).

Conversely, suppose that you have a word $w \in L$: it can be broken down as $x y^r z y x^r$ with $|x| \ge 1$ and $|y| \ge 1$. If $|x| \ge 2$ then write $x = a b x'$: we have $w = a b x' y^r z y x'^r b a = a b z' b a$ where $z' = x' y^r z y x'^r$. If $|x| = 1$ then write $y = b y'$ and do a similar decomposition.

Conclusion: another way of describing $L$ is $L = \{a b z b a \mid a,b \in \{0,1\}, z \in \{0,1\}^{+}\}$. This language is regular; a regular expression that matches it is
00..*11|01..*10|10..*01|11..*11. Intuitively, an autotomaton recognizing $L$ only needs a finite amount of memory, because it only needs to memorize the first two letters.