Prove $\neg (p \land q) \vdash \neg p \lor \neg q$ by natural deduction

Fairly close.   Always keep an eye on the prize.   You want to be able to contradict the first premise, $\lnot(p\land q)$, so that requires deriving $p$ and $q$, then introducing a conjunction.

Thus the purpose of those subproof are to be proofs by contradiction too.

$$\def\bot{\mathcal F}\def\fitch#1#2{\quad\begin{array}{|l} #1\\\hline #2\end{array}} \fitch{~~1.~\lnot(p\land q)}{\fitch{~~2.~\lnot(\lnot p\lor\lnot q)}{\\\fitch{~~3.~\lnot p}{~~4.~\lnot p\lor \lnot q\hspace{4ex}\lor\mathsf I, 3\\~~5.~\bot\hspace{9ex}\lnot\mathsf E,2,4}\\~~6.~p\hspace{13ex}\mathsf {PBC},3{-}5\\\\\fitch{~~7.~\lnot q}{~~8.~\lnot p\lor \lnot q\hspace{4ex}\lor\mathsf I, 7\\~~9.~\bot\hspace{9ex}\lnot\mathsf E,2,8}\\10.~q\hspace{13ex}\mathsf {PBC},7{-}9\\\\11.~p\land q\hspace{9ex}\land\mathsf I, 6,10\\12.~\bot\hspace{12ex}\lnot\mathsf E,1,11}\\13.~~\lnot p\lor\lnot q\hspace{9ex}\mathsf {PBC},2{-}12}$$

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PS: The PBC steps contain the LEM, in the form of double negation elimination (DNE). $$\begin{array}{|l}\fitch{~~3.~\lnot p}{~~4.~\lnot p\lor \lnot q\hspace{4ex}\lor\mathsf I, 3\\~~5.~\bot\hspace{9ex}\lnot\mathsf E,2,4}\\~~6.1.~~\lnot\lnot p\hspace{10ex}\neg\mathsf I, 3{-}5\\~~6.2.~p\hspace{13ex}\lnot\lnot\,\mathsf E,6.1\end{array}$$