2NF: Remove Partial Dependencies
R = {A, B, C, D, E, F, G, H, I, J}
includes partial dependencies.
D and E depend only on A, F depends only on B, G, H, I and J don't depend on the key (directly) at all.
R0 = {A, B, C}
R1 = {A, D, E, I, J}
R2 = {B, F, G, H}
R0, R1, and R2 contain no partial dependencies (or repeating groups) so they are 2NF. However R1 and R2 are still an issue, because they contain transitive dependencies.
3NF: Remove Transitive Dependencies
I and J depend on D, not on the key of R1. Therefore you need to further normalize R1 as follows:
R1 = {A, D, E, I, J}
R1a = {A, D, E}
R1b = {D, I, J}
Similarly, G and H depend only on F so R2 must be decomposed as follows:
R2 = {B, F, G, H}
R2a = {B, F}
R2b = {F, G, H}
Now all of your remaining relations (R0, R1a, R1b, R2a, R2b) are devoid of repeating groups, partial dependencies and transitive dependencies. That means your relations are in 3NF.
When you are looking at an relation that hasn't been normalized and a series of dependencies, you can often normalize by inspection just by recognizing what your primary keys are going to be. Any attribute or combination of attributes that functionally determine other attributes are going to end up as primary keys. Once you've got your primary keys defined, you just need to figure out which non-key attributes go with each key. This is obvious from the statement of what your functional dependencies are.
Best Answer
You can't tell the functional dependencies just from the schema.
If you knew that the schema was in third normal form (3NF) or higher and you knew that there are no other candidate keys other than the primary key, then you would know that every non-key attribute was fully functionally dependent on the primary key.
Except for this particular situation, there is nothing about a schema that conveys functional dependency. The purpose of a schema is to convey table structure, not functional dependencies.