We'll first apply, to the right-hand side of your biconditional, the following identity:
$$P \Rightarrow Q \equiv \lnot P \lor Q\quad \tag{logically equivalent}$$
Each side of the above equivalence implies the other (they are logically equivalent).
Applying this to the right had side of where you left off $(1)$, we get $(2)$:
$$((P\wedge Q)\vee (\neg P \wedge \neg Q))\iff((P\Rightarrow Q)\wedge (Q\Rightarrow P))\tag{1}$$
$$((P\wedge Q)\vee (\neg P \wedge \neg Q))\iff(\lnot P \lor Q)\wedge (\lnot Q \lor P))\tag{2}$$
Try using distribution on the right hand side (twice) until you arrive at an equivalent expression as the left hand side of the biconditional. Putting this all together, we get:
$$\begin{align} ((P\land Q) \lor (\lnot P \land \lnot Q))
& \iff ((P \Rightarrow Q) \land (Q \Rightarrow P)) \\ \\
& \iff ((\lnot P \lor Q) \land (\lnot Q \lor P)) \\ \\
& \iff [\lnot P \land (\lnot Q \lor P)] \lor [Q \land (\lnot Q \lor P)] \\ \\
& \iff [(\lnot P \land \lnot Q) \lor (\lnot P \land P)] \lor [(Q\land \lnot Q) \lor (Q \land P)] \\ \\
& \iff (\lnot P \land \lnot Q) \lor F \lor F \lor (Q \land P) \\ \\
& \iff (\lnot P \land \lnot Q) \lor (Q\land P) \\ \\
& \iff ((P \land Q) \lor (\lnot P \land \lnot Q))
\end{align}$$
By using the given identity, applying the distributive law twice, and remembering that $A \land \lnot A \equiv F\;$ and $\;A \lor F \equiv A$, we have obtained, from the right hand side of our biconditional, the equivalent expression on the left-hand side of $(2)$.
You should start supposing the contrary of the conclusion, that is,
$$\neg(\neg p \vee (\neg q\vee r)))$$
Now, apply twice the De Morgan's law for $\neg(X\vee Y)$:
$$p\wedge q \wedge\neg r$$
Eliminate $\wedge$ to obtain $\neg r$. Now, use the premise and eliminate $\vee$:
$$\neg(\neg p\vee q)$$
Apply again De Morgan's law:
$$p\wedge \neg q$$
Now, eleminate $\wedge$ in two previous formulae to get $q$ and $\neg q$. Insert $\wedge$ and you have the contradiction.
Best Answer
You apparently have a rule that allows you to go back and forth between $A\Rightarrow B$ and $\lnot A\lor B$. You’ve used it in the forward direction to go from $P\Rightarrow Q$ to $\lnot P\lor Q$, with $A=P$ and $B=Q$. Now use it in the reverse direction to go from $\lnot\lnot Q\lor\lnot P$ to $\lnot Q\Rightarrow \lnot P$ by taking $A=\lnot Q$ and $B=\lnot P$.