Is the language of all words of the form $wxw^Rx^R$ context-free?
Denote your language by $L$, and consider
$$
L' = L \cap 10^*1110^*10^*10^+11.
$$
Any word in $L'$ is of the form
$$
10^a1110^b10^c10^d11,
$$
where $d &gt; 0$.
Also, it is of the form $wxw^Rx^R$. Since both $wx$ and $w^Rx^R$ contain the same number of $1$'s, necessarily
\begin{align}
wx &amp;= 10^a1110^{b_1} \\
w^Rx^R &amp;= 0^{b_2}10^c10^d11
\end{align}
where $b_1 + b_2 = b$. We see that $x^R$ ends with $110$ (since $d&gt;0$), and so $x$ starts with $110$. Hence $w = 10^a1$ and $x=110^{b_1}$, implying that $w^Rx^R = 10^a10^{b_1}11$. This shows that $b_2 = 0$ hence $b_1 = b$, and furthermore $c = a$ and $d = b$. That is, the word is of the form
$$
10^a1110^b10^a10^b11,
$$
where $b &gt; 0$. Conversely, every word of this form is in $L'$ (take $w = 10^a1$ and $x = 110^b$).
Let $h\colon \{1,2,3,4,5\} \to \{0,1\}$ be the homomorphism given by $h(1) = 1$ and $h(2) = h(3) = h(4) = h(5) = 0$, and let $d\colon \{1,2,3,4,5\} \to \{2,3,4,5\}$ be the homomorphism that deletes $1$. Then
$$
d(h^{-1}(L') \cap 12^*1113^*14^*15^*11) \cup \{2^n4^n : n \geq 0\} = \{2^n3^m4^n5^m : n,m \geq 0\}.
$$
The latter language is well-known not to be context-free. Since all our operations preserve being context-free, we conclude that $L$ is also not context-free.