Proof of composition of one way function is not one way in general
"for a given output of g(x,y) = 0|x| of length l (say)," "the output of g(any x of length l, 0l)"

is "the same as the earlier output", which means that is "a possible preimage of 0|x|".

Accordingly, for ​ $g\hspace{-0.04 in}\circ \hspace{-0.03 in}f$ ​ to be one-way, it had better be the case that

$\operatorname{Prob}_{\langle \hspace{.02 in}x,y\rangle \hspace{.02 in}\leftarrow \hspace{.02 in}\left(\hspace{-0.03 in}\{\hspace{-0.02 in}0,1\hspace{-0.02 in}\}^{\hspace{.02 in}L}\hspace{-0.03 in}\right)^{\hspace{.05 in}2}}\hspace{-0.1 in}\left[g(x,\hspace{-0.04 in}y) = 0^{\hspace{.03 in}L}\hspace{-0.0 in}\right] \;\;\;$ is negligible.