Operation on two CFLs which results in a non-CFL
Let's denote your operation by $\diamond$, so the question is, if $L_1$ and $L_2$ are CFLs, is $L_1\diamond L_2$ always a CFL? The answer is no.

Let $L_1=L_2=\{0^n1^n\mid n\ge 0\}$. This is, of course, a CFL. However, 
$$
L_1\diamond L_2=\{0^n1^n0^n1^n\mid n\ge 0\}
$$
and it's not too hard to show that this language is not context-free.