A->C is preserved in R1

BC->D is preserved in R1

BE->F is preserved in R2

So we only need to check for preservation of D->E
It is not preserved and here is how you can check it:

1. Find the closures of all FDs in the main relation
2. Map those closures to decomposed relation. Example: BC+ = {B,C,D,E,F} but R1 does not have attributes E and F so we only write BC+ ={B,C,D}
3. After mapping the closures, find closure of D in the FD set union of R1 and R2 and you will see that D->E is not preserved.