Use equivalence properties to show that $[\neg(p\leftrightarrow q)] \iff [(p\leftrightarrow (\neg q))]$ I've really been struggling to get the correct answer on this one. I've tried several ways but this is best I've been able to get it down to so far:
$[\neg(p\leftrightarrow q)] \iff [(p\leftrightarrow (\neg q))]$
(i)$[\neg(p\leftrightarrow q)]\iff \neg[(p\to q)\land (q\to p)]$
(ii)$[\neg(p\leftrightarrow q)]\iff \neg(p\to q) \lor \neg(\neg q\lor p)$
(iii)$[\neg(p\leftrightarrow q)]\iff ((p\land \neg q)\lor q) \land p$
(iv)$(q\lor p) \land (q\lor \neg q) \land \neg p$
$q \lor p \land \neg p \land$ 0
Note: 0 means contradiction
 A: One way to do it is the following:
$$\neg(p\leftrightarrow q)\leftrightarrow(p\leftrightarrow\neg q)$$
$$\Longleftrightarrow\Big(\neg(p\leftrightarrow q)\land(p\leftrightarrow\neg q)\Big)\lor\Big((p\leftrightarrow q)\land\neg(p\leftrightarrow\neg q)\Big)$$
$$\Longleftrightarrow\bigg(\Big(\neg p\land q)\lor(p\land\neg q)\Big)\land\Big((p\land\neg q)\lor(\neg p\land q)\Big)\bigg)\lor\bigg(\Big((\neg p\land\neg q)\lor(p\land q)\Big)\land\Big((\neg p\land\neg q)\lor(p\land q)\Big)\bigg)$$
$$\Longleftrightarrow (\neg p\land q)\lor(p\land \neg q)\lor(\neg p\land\neg q)\lor(p\land q)$$
First, notice that to go from the second $\Longleftrightarrow$ to the third, we've contracted the conjunctions - since $(A\land A)\leftrightarrow A$.
Now, we're left with the last line, which is essentially a disjunction of all possible truth values assigned to $p$ and $q$ - take a moment to convince yourself that it is indeed the case. Therefore, since we've exhausted all possible cases, the proposition will always be true, i.e. we have shown that $\neg(p\leftrightarrow q)$ and $(p\leftrightarrow\neg q)$ are equivalent.
Finally, you may also want to try the following technique: start by negating the statement you wish to prove is a tautology, and work you way down to the truth values of each sentence letter (either by truth tables or tableaux); if the negated proposition can never be true (i.e. if it is a logical falsehood), then by negation your original proposition must be always true (i.e. a tautology).

EDIT: after paying more attention to your question, I'll give you an answer that might be more along the lines of what you first tried to do.
\begin{align}\neg(p\leftrightarrow q)\Longleftrightarrow&\neg\big((p\to q)\land(q\to p)\big)\\\Longleftrightarrow&\neg(p\to q)\lor\neg(q\to p)\\\Longleftrightarrow&(p\land\neg q)\lor(q\land\neg p)\\\Longleftrightarrow&(p\leftrightarrow\neg q)\end{align}
where to move from the penultimate to the last line, we have simply used the definition of the biconditional ("equivalence").
