If $N$ is a normal subgroup of $G$, show $Z(G)N/N \subset Z(G/N)$ [closed]

You don't need to use that equality to get the result. If $gnN$ is an element of $Z(G)N/N$, where $g \in Z(G)$ and $n \in N$, then for any element $xN \in G/N$, where $x \in G$, you have

$$(gnN)(xN) = (gN)(xN) = gxN = xgN = (xN)(gN) = (xN)(gnN).$$

This shows that $gnN$ commutes with an arbitrary element of $G/N$, so it lies in the center $Z(G/N)$ of $G/N$.

Remark: If $X$ is any subset of $G$, then $X/N$ is by definition the set of cosets of the form $xN$, where $x \in N$. Clearly $Z(G)N/N = Z(G)/N$. So a typical element of $Z(G)N/N$ is just $gN$, for $g \in Z(G)$.

Additional remark: equality in $Z(G)N/N \subseteq Z(G/N)$ does not have to occur, take for example $G=Q$, the quaternion group of order $8$ and $N=Z(G)$. Then $Z(G)N/N=\overline{1}$, and $Z(G/N) \cong C_2 \times C_2$.

However, in general, if $N \cap G'=1$, then $Z(G/N)=Z(G)N/N=Z(G)/N$: if $g \in G, n \in N$, then $[n,g]=n^{-1}(g^{-1}ng) \in N \cap G'$. So $N \subseteq Z(G)$ in this case. If $Z(G/N)=K/N$, then $[K,G] \subseteq N$, but also $[K,G] \subseteq [G,G]$, hence $[K,G]=1$, that is $K \subseteq Z(G)$. So $Z(G/N) \subseteq Z(G)/N$ and the reverse inclusion was already proven by @D_S.