1 1) F --> [ ( C --> C ) --> G ] Premise
2 2) G --> { [ H --> ( E --> H ) ] --> ( K * ~ K ) } Premise
3 3) F Assumption
1,3 4) ( C --> C ) --> G 1,3, Modus Ponens (MP)
5 5) C Assumption
6) C->C 5, Conditional Introduction (CI)
1,3 7) G 4,6 MP
1,2,3 8) [ H --> ( E --> H ) ] --> ( K * ~ K ) 2, 7 MP
9 9) E Assumption
10 10) H Assumption
10 11) E-->H 9,10 CI
12) H --> (E-->H) 10,11 CI
1,2,3 13) K & -K 8,12 MP
1,2 14) -F 3,13 Law of Non Contradiction

That gives you an indirect proof of -F given the two premises.

If by conditional proof, you mean simply prove that if premises 1 and 2 hold, then -F follows, then you can presumably just add the following steps
1,2 15) (F --> [ ( C --> C ) --> G ]) & (G --> { [ H --> ( E --> H ) ] --> ( K * ~ K ) } ) 1,2 Conjunction Int
16) [ (F --> [ ( C --> C ) --> G ]) & (G --> { [ H --> ( E --> H ) ] --> ( K * ~ K ) } )] --> -F 14,15 CI