How to derive ~(P → Q) → P?

the book only gives "modus tollens", "modus ponens", "douple negation", "repetion" (A, therefore A), direct derivation, conditional derivation, indirect derivation and subderivation.

You are asked to derive a conditional statement, $\neg (P\to Q)~\to~ P$ so use a conditional derivation.   So make a subderivation of the consequent, $P$, under the assumption of the antecedent, $\neg(P\to Q)$.

You have no way to directly derive $P$ under that assumption, so that calls for an indirect derivation. Thus make a subderivation of a contradiction under the assumption of the negation, $\neg P$.

Now derive a contraduction under the assumptions $\neg P$ and $\neg(P\to Q)$ will involve either deriving $P$ or $P\to Q$. Since if you could directly derive $P$ you would not be here... You are seeking to derive the conditional statement $P\to Q$, so ...

$$\def\fitch#1#2{~~\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{}{\fitch{\neg (P\to Q)}{\fitch{\neg P}{~~\vdots\\P\to Q\\\bot}\\P}\\\neg(P\to Q)\to P}$$

this is a derivation using natural deduction rules from the textbook Gamut L.T.F. "Logic, language and meaning". It may help you with your derivation.

1. ¬(P→Q) - assumption
2. ¬P - assumption
3. P - assumption
4. ⊥ in 3,4
5. Q - EFSQ (explosion principle) (close assumption 3)
6. P→Q - Rule Introduction of → in 3,5
7. ⊥ in 1,6 (close assumption 2)
8. ¬¬P - Rule Introduction of ¬ in 2
9. P - Rule Elimination of ¬¬ in 8 (close assumption 1)
10. ¬(P→Q)→P - Rule Introduction of → in 1,9