Is the following deduction valid?
Consider $$\lnot X \land ((Y \land \lnot Z) \lor (\lnot Y \land Z)).$$
Manipulating via distributive laws:
\begin{align*} &= [\lnot X \lor (Y \land \lnot Z)] \land [\lnot X \lor (\lnot Y \land Z)] \\ &= [(\lnot X \land Y) \lor (\lnot X \land \lnot Z)] \land [(\lnot X \land \lnot Y) \lor (\lnot X \lor Z)] \\ &= {[[{(\lnot X \land Y) \lor (\lnot X \land \lnot Z)] \lor (\lnot X \land \lnot Y)}] \land [{[(\lnot X \land Y) \lor (\lnot X \land \lnot Z)] \lor (\lnot X \lor Z)}]} \\ &= {[{[(\lnot X \land Y) \lor (\lnot X \land \lnot Y)] \lor (\lnot X \land \lnot Z)]} \land [{[(\lnot X \land Z) \lor (\lnot X \land \lnot Z)] \lor (\lnot X \land Y)}]} \\ &= [\lnot X \lor (\lnot X \land \lnot Z)] \land [\lnot X \lor (\lnot X \land Y)] \\ &= \lnot X \land \lnot X \\ &= \lnot X. \end{align*}
You misused De Morgan's Laws. Using distributive laws, it will simply be just:
\begin{align*} &= \lnot X \land ((Y \land \lnot Z) \lor (\lnot Y \land Z)) \\ &= (\lnot X \land Y \land \lnot Z) \lor (\lnot X \land \lnot Y \land Z) \end{align*} That will be the most simple form of your problem, I believe.