Suppose that $ G/Z(G) $ is perfect. Can we conclude that the commutator subgroup $ [G,G] $ is perfect?
I think that $ [G,G] $ must be a perfect central extension of $ G/Z(G) $ and that $ G $ has the structure of a central product of $ Z(G) $ with $ [G,G] $ over $ Z([G,G]) $ $$ G\cong Z(G) \circ_{Z([G,G])} [G,G] $$
Also I think $ [G,G] $ is the maximal perfect subgroup of $ G $.
In general, if $N \unlhd G$, then $(G/N)'=G'N/N$, so if $G/Z(G)$ is non-abelian simple, it follows that $(G/Z(G))'=G'Z(G)/Z(G)=G/Z(G)$. This means that $G=G'Z(G)$ and this yields (see also the remarks of Derek Holt), $G'=G''$, so $G'$ is perfect. In addition, $Z_2(G)=Z(G)$, which I leave as an exercise to you.