In your approach, you made a mistake in the derivation on the third line, it should be:
$$[\lnot (\lnot p \lor q) \lor \lnot (q \rightarrow r)] \lor (p \rightarrow r)$$
so that your fourth line becomes:
$$[(p\land \lnot q) \lor (q \land \lnot r)] \lor (p \rightarrow r)$$
At this point, there isn't really much to be done except for a tedious case-by-case argument ("Suppose $p$ is true. [...] Now suppose $p$ is false [...].").
Let us take a step back. When proving a tautology (which is not easily seen to be so by virtue of known theorems), it is often a more productive technique to try and prove that its negation is a contradiction. The idea behind this is that if the original $\phi$ is to be a tautology, then by investigating its negation $\neg \phi$, we will obtain stricter and stricter conditions on $p, q, r$ to allow $\neg \phi$ to be true, until we get to the conclusion that there are no possibilities left, i.e. that $\phi$ is a tautology.
Essentially, one converts a universal check ("All interpretations yield true") to a search for an example ("Some interpretation does not yield true") -- keeping in mind that we want the latter search to be fruitless. Basically, it's easier to find the black sheep in a flock than to count the white ones. For a more detailed account of this approach, you can search for terms like semantic tableaux or analytic tableaux.
Returning to our concrete example, let us see how this approach works out. We use repeatedly the equivalences $p \to q \iff \neg p \lor q$ and $\neg (p \to q) \iff p \land \neg q$:
\begin{align*}
& \neg\big([(p\rightarrow q) \land (q \rightarrow r)] \rightarrow (p \rightarrow r )\big)\\
\iff& (p\to q)\land (q\to r) \land \neg(p \to r) \\
\iff& (\neg p \lor q) \land (\neg q \lor r) \land p \land \neg r
\end{align*}
From the last two clauses, we see that $p$ must be true and $r$ must be false. But then for $\neg p \lor q$ to be true, $q$ must be true as well. Similarly, $\neg q$ must be true in order for $\neg q \lor r$ to be true as well. Hence we need $q$ to be both true and false at the same time, a contradiction.
We conclude that $[(p\rightarrow q) \land (q \rightarrow r)] \rightarrow (p \rightarrow r )$ is a tautology, as desired.
Resolution is not an equivalence.
The easiest way to derive it is to use the tautological equivalence between : $A \to B$ and $\lnot A \lor B$.
Thus :
$(p∨q)∧(¬p∨r)$
is equivalent to :
$(\lnot q \to p) \land (p \to r).$
Then, applying Hypothetical syllogism, we get :
$\lnot q \to r$
i.e. : $q \lor r$.
Best Answer
In general, in propositional classical logic (which is the logic where truth tables make sense), a standard way to prove that a formula is a tautology without using truth table is:
This is sufficient since, thanks to the soundness theorem, every formula that is provable in such a system is a tautology.
More specifically, concerning your question, a derivation of $\big( (p \lor q) \land ((p \to r) \land (q \to r)) \big) \to r$ in propositional classical natural deduction is the following:
Thus, by the soundness theorem for propositional classical natural deduction, the formula $\big( (p \lor q) \land ((p \to r) \land (q \to r)) \big) \to r$ is a tautology.
(Remark: Actually, the derivation above does not use the rule RAA (reductio ad absurdum), hence the formula is provable in intuitionistic propositional logic, which is weaker that propositional classical logic.)
Another way (equivalent to the previous one and closer to your approach) to prove that $\big( (p \lor q) \land ((p \to r) \land (q \to r)) \big) \to r$ is a tautology is the following argument, where $\equiv$ stands for logical equivalence, and $\top$ for truth (or tautology):
\begin{align} &\quad \big( (p \lor q) \land (p \to r) \land (q \to r) \big) \to r \\ &\equiv \lnot \big( (p \lor q) \land (p \to r) \land (q \to r) \big) \lor r \\ &\equiv \big( \lnot(p \lor q) \lor \lnot(p \to r) \lor \lnot(q \to r) \big) \lor r \\ &\equiv (\lnot p \land \lnot q) \lor (p \land \lnot r) \lor (q \land \lnot r) \lor r \\ &\equiv (\lnot p \land \lnot q) \lor (p \land \lnot r) \lor ((q \lor r) \land (\lnot r \lor r)) \\ &\equiv (\lnot p \land \lnot q) \lor (p \land \lnot r) \lor ((q \lor r) \land \top) \\ &\equiv (\lnot p \land \lnot q) \lor (p \land \lnot r) \lor q \lor r \\ &\equiv ((\lnot p \land \lnot q) \lor q) \lor ((p \land \lnot r) \lor r) \\ &\equiv ((\lnot p \lor q) \land (\lnot q \lor q)) \lor ((p \lor r) \land (\lnot r \lor r)) \\ &\equiv ((\lnot p \lor q) \land \top) \lor ((p \lor r) \land \top) \\ &\equiv \lnot p \lor q \lor p \lor r \\ &\equiv (\lnot p \lor p) \lor q \lor r \\ &\equiv \top \lor q \lor r \\ &\equiv \top \end{align}