Prove or disprove L is regular
Your grammar generates the following language:
$$
L = \{ w0^*1^*0^*1^*\bar{w}^R : w \in \{0,1\}^* \},
$$
where $\bar{w}$ is obtained from $w$ by switching 0s and 1s. Consider now $L' = L \cap 0^*1010101^*$. It is not hard to check that $L$ contains all words of the form $0^n1010101^n$ (take $w = 0^n1$ and $0101 \in 0^*1^*0^*1^*$). Conversely, if $0^n1010101^m \in L'$ and $n \neq m$, say $n &gt; m$, then the decomposition $0^n1010101^m = wx\bar{w}^R$, where $x \in 0^*1^*0^*1^*$, must satisfy that $\bar{w}^R = 1^k$ for some $k \leq m$; but then $x = 0^{n-k}1010101^{m-k} \notin 0^*1^*0^*1^*$. We conclude that
$$
L' := L \cap 0^*1010101^* = \{ 0^n1010101^n : n \geq 0 \}.
$$
It is not hard to check that $L'$ is not regular (for example, the words $0^n$ are pairwise inequivalent), and so $L$ also isn't regular.