R(ABCDEFH) { E → F, C → DEH}
here, Candidate key will be {ABC}. So, C –> DEH is a partial functional dependency. So, its not in 2NF.
2NF decomposition will be: R1(CDEHF), R2(ABC)
Also, since its not in 2NF. so its also not in 3NF.
For 3NF decomposition will be: R1(EF), R2(CDEH), R3(ABC)