 Solution

False. Use only the directed edge $u\to v$, but choose $v$ first in the outer loop of depth-first search. It finishes before a second discovery tree starts at $u$: $d[v]=1$, $f[v]=2$, $d[u]=3$, $f[u]=4$. The edge supplies the required directed path and $f[u]>f[v]$, yet $d[u]>d[v]$. A path to an already black vertex need not put the two vertices into one discovery tree.