Applying second and third Normal Form
The only candidate key of your relation is {D DA HA L NF} (perhaps with R you mean D?)

You can verify this by calculating the closure of those attributes, {D DA HA L NF}+, and seeing that it contains all the attributes, while, if you remove any one of them, the closure of the remaining set does not contains all the attributes (this is the definition of a candidate key).

The relation is only in first normal form, since the second normal form requires the absence of partial dependencies, that is of dependencies in which non-prime attributes (i.e. attributes not belonging to any key) depends only on part of a key. In this case only the last functional dependency is not partial.