In DFS , edge will go in longest depth. So, as edge must will go from u to v , so, DFS will include , but not necessarily v to u .
Here d(u) actually the total time taken from starting vertex say S to vertex u
Now 3 pictures clears all doubt.
Here d(u)=2, d(v)=5
f(u) will go till last depth of this path starting from u . So, it will be 3+2=5
f(v)=2
So, option B) and C) eliminates
In second picture
d(u)=2
d(v)=2
f(u)=6
f(v)=3 (As, uv always exists in the DFS)
It eliminates option A)
In 3rd picture, if there is a loop
Here, f(u)=6,
f(v)=3 (As, f(u) already taken uv, f(v) neednot to take uv again)
So, after all , we can conclude only option D) will remain as always true