is $0^x1^y$ context-free?
Your first language is actually regular: if $s$ is the substitution mapping $0$ to $\{0,1\}$, then
$$
\{ 0^x 1^y : 0^{x+y} \in L \} = s(L \cap 0^*) \cap 0^*1^*.
$$
Your second language is context-free, since we can write it as $L \cap 0^*$ concatenated with $\{0^y 1^y : y \geq 0\}$. It need not be regular, as the example of $L = \{\epsilon\}$ demonstrates: in this case, your language is just $\{0^y1^y : y \geq 0\}$.