Okay, so we have Modus Ponens, Modus Tollens, and Double-Negation Elimination $$ \frac{p\to q \qquad p}{q}\text{MP} \qquad \frac{p\to q\qquad \neg q}{\neg p}\text{MT} \qquad \frac{\neg \neg p}{p}\text{DNE} $$ plus natural deduction: $$ \begin{array}{c}p \\ \vdots \\ \frac{q}{p\to q} \end{array}$$
We want eventually to conclude $\neg(p\to q)\to p$. The natural way to conclude an implication is by deduction, so $$ \tag{1} \textbf{Assume: } \neg(p\to q) $$
How are we going to get from there to $p$? The only way to pick apart a singly-negated premise with the tools we're given is to use MT, and for that we'd need to have something of the form $X \to (p\to q)$. We'd then get the $X$ in negated form, which does not look promising unless $X$ is already negated once, in which case we can use DNE on it. So it might be useful to see if we can produce $\neg p \to (p\to q)$. A bit of truth-tabling tells us that this indeed is a tautology, which is reassuring. But how to prove that? Again, the deduction theorem: $$ \tag{2} \textbf{Assume: } \neg p $$ $$ \tag{3} \textbf{Assume: } p $$
And we now want to conclude $q$ from $\neg p$ and $p$. Again the only way to make use of the $\neg p$ assumption is to use MT on it, so we need to have something of the form $Y\to p$ first. But we can get that for any $Y$ because we already have $p$, so we can just choose $Y$ with a view to being able to prove $q$ from $\neg Y$ -- i.e., let $Y$ be $\neg q$ and do
$$ \tag{4} \textbf{Assume: } \neg q $$ $$ \tag{5} \textbf{By assumption }(3): p $$ $$ \tag{6} \textbf{Discharge assumption }(4): \neg q \to p$$ $$ \tag{7} \text{MT}(4,2): \neg \neg q$$ $$ \tag{8} \text{DNE}(7): q$$
And now we just have to carry out the rest of the plans we already made: $$ \tag{9} \textbf{Discharge assumption }(3): p\to q$$ $$ \tag{10} \textbf{Discharge assumption }(2): \neg p \to (p \to q)$$ $$ \tag{11} \text{MT}(10,1): \neg \neg p $$ $$ \tag{12} \text{DNE}(11): p$$ $$ \tag{13} \textbf{Discharge assumption }(1): \neg(p\to q)\to p$$