What we would like to prove is a conjunction, so it suffices to prove each conjunct separately and then glue them together at the end. This problem would probably be easier and more intuitive using proof by contradiction, but after talking with the asker, I will provide a direct proof.
$$\begin{array}{lr}
1. & (P \rightarrow Q)\wedge(Q \rightarrow R) & \text{Premise} \\
2. & P \rightarrow Q &\text{Simplification, 1}\\
3. & Q \rightarrow R & \text{Simplification, 1}\\
4. & \neg{P} \vee Q & \text{Conditional Law, 2}\\
5. & \neg{Q} \vee R & \text{Conditional Law, 3}\\
6. & Q \vee \neg{Q} & \text{Tautology} \\
7. & \neg{P} \vee R & \text{Constructive Dilemma, 4,5,6}\\
8. & P \rightarrow R & \text{Conditional Law, 7}\\
9. & (P \rightarrow Q) \vee (Q \rightarrow R) &\text{Addition, 2}\\
10. & (Q \rightarrow P) \vee (Q \rightarrow R) &\text{Addition, 3}\\
11. & (P \rightarrow Q) \vee (R \rightarrow Q) &\text{Addition, 2}\\
12. & Q \vee \neg{Q} & \text{Tautology}\\
13. & (Q \vee \neg{Q}) \vee (P \vee \neg{R}) & \text{Addidition, 12}\\
14. & (\neg{Q} \vee P) \vee (\neg{R} \vee Q) & \text{Associative Law, 13}\\
15. & (Q \rightarrow P) \vee (R \rightarrow Q) & \text{Conditional Law, 14}\\
16. & \big((P \rightarrow Q) \vee (Q \rightarrow R)\big)\wedge \big((Q \rightarrow P) \vee (Q \rightarrow R)\big) & \text{Conjunction, 9,10}\\
17. & \big((P \rightarrow Q) \vee (R \rightarrow Q)\big)\wedge \big((Q \rightarrow P) \vee (R \rightarrow Q)\big) & \text{Conjunction, 11,15}\\
18. & \big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (Q \rightarrow R) & \text{Distributive Law, 16}\\
19. & \big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (R \rightarrow Q) & \text{Distributive Law, 17}\\
20. & \Big(\big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (Q \rightarrow R)\Big) \wedge & \\
&\Big(\big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (R \rightarrow Q)\Big) & \text{Conjunction, 18,19}\\
21. & \big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee \big((Q \rightarrow R) \wedge (R \rightarrow Q) \big) & \text{Distributive Law, 20}\\
22. & (P \equiv Q) \vee (Q \equiv R) & \text{Definition of Biconditional, 21}\\
\therefore & (P \rightarrow R)\wedge \big((P \equiv Q) \vee (Q \equiv R)\big) & \text{Conjunction, 8,22}
\end{array}$$
As desired.
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
Okay, I firstly did the truth table ( I didn't do it first because I wanted to work with the different laws and learn to see how to apply them). This is left as it follows:
\begin{array}{ccccc} p & q & (\neg p \lor \neg q) & (p \lor q) & (\neg p \lor \neg q) \land (p \lor q)\\ \hline \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 \\ \end{array}
After this, and seeing that this was equal to $ \neg(p \leftrightarrow q) $ I tried to find it using the different laws. The procedure I followed was:
$$(\neg p \lor \neg q) \land (p \lor q) = [ (\neg p \lor \neg q) \land p] \lor[(\neg p \lor \neg q) \land q] \\ = [(\neg q \land p) \lor (\neg p \land q)] \\ = [\neg(q \lor \neg p) \lor \neg (p \lor \neg q)] \\ = [\neg (p \rightarrow q) \lor \neg(q \rightarrow p)] \\ = \neg [(p \rightarrow q) \land (q \rightarrow p)] \\ = \neg (p \leftrightarrow q)$$