I can give it a try!
I'll assume you are using the natural deduction system that has the normal introduction and elimination rules for the logical symbols, the so-called "intuitionistic fragment" of propositional logic, together with double negation elimination "from $\lnot \lnot A$, conclude $A$'' (DNE). If you started out by adding the rule "if under assumption $\lnot p$ we have $\bot$, then $p$'', we can get (DNE) also.
Open the following assumptions in this order: 1) $(E \rightarrow F) \rightarrow E$, 2) $\lnot E$ and 3) $E \rightarrow F$. From 1) and 2), conclude $E$ contradicting 2), so that we can close assumption 3) and conclude $\lnot (E \rightarrow F)$ by the I-$\lnot$ rule. From this we get $\lnot \lnot E$ and then $E$. This contradicts 2) again, so we can close 2) and get $\lnot \lnot E$ by I-$\lnot$ rule again. By DNL, we have $E$. Hence we can close assumption 1) to get $((E\rightarrow F) \rightarrow E) \rightarrow E$ by I-$\rightarrow$
We used classical contradiction when we closed off 2) to conclude $E$ by showing that $\lnot E$ is absurd, and also to get from $\lnot \lnot E$ to $E$.
If the order of which assumptions to open seems ad-hoc, it's basically dictated by what you're trying to prove. Since Pierce's law has a $\rightarrow$ as an outer connective, it's prudent to try using the I-$\rightarrow$ first. Then we are left to show $E$ from assumption 1), which is handily proved by DNL.
Hope that helps!