Without constructing a truth table show that the statement formula ~(~p→~q)→~(q→p) is a tautology

Remember that $p \vee \neg p$ is a tautology, meaning that the statement always evaluates to true. Also, $p \rightarrow q \equiv \neg p \vee q$.

So we have \begin{align} \neg (\neg p \rightarrow \neg q) \rightarrow \neg (q \rightarrow p) &\equiv \neg \left[\neg(\neg p) \vee \neg q\right] \rightarrow \neg (\neg q \vee p) & \text{(definition of implication (x2) )}\\ &\equiv \neg \left(p \vee \neg q\right) \rightarrow \neg (\neg q \vee p) \\ &\equiv \neg \left(p \vee \neg q\right) \rightarrow \neg (p \vee \neg q) & \text{(rearranging terms)}\\ &\equiv \neg \left[\neg (p \vee \neg q) \right] \vee \neg (p \vee \neg q) & \text{(definition of implication))} \\ &\equiv (p \vee \neg q) \vee \neg ( p \vee \neg q) \end{align}

Now let $r = p \vee \neg q$. We then have $(p \vee \neg q) \vee \neg (p \vee \neg q) \equiv r \vee \neg r$, which is a tautology.