Warning: (YCor) the following argument is mistaken as was pointed out by Derek Holt: the assertion that the abelianization of a torsion-free nilpotent group is torsion-free is hopelessly wrong.
The answer is "yes" because every f.g. nilpotent group has a torsion-free finite index subgroup and because the abelianization of a torsion-free nilpotent group is torsion-free. I assume that by "rank" you meant the torsion-free (${\mathbb Q}$-)rank.
This group is not finitely presentable. Indeed, write $F$ for the given free group and $F'$ for its derived subgroup. The map
$$F\ltimes F'\to F\times F,\quad (f,g)\mapsto (f,fg)$$
is an injective group homomorphism, and its image is the fibre product of two copies of $F$ over the abelianization map $F\to F/F'$. (Alternatively, start from this fibre product, and observe that it is the semidirect product $H\ltimes K$, where $H\simeq F$ is the diagonal and $K=F'\times\{1\}$.)
It is a result of Baumslag and Roseblade that a subgroup of a direct product of two free groups is "almost never" finitely presented, i.e. not finitely presentable unless one of the projection is injective on the subgroup, or if it is virtually the direct product of intersection with the factors. Since this is not such an exception, the above group is not finitely presentable.
Reference: G. Baumslag and J. E. Roseblade, Subgroups of direct products of free groups, J. London Math. Soc. (2) 30 (1984), 44-52. link DOI behind paywall
PS: the same argument works equally if $N$ is an arbitrary nontrivial normal subgroup of infinite index in $F$: then $F\ltimes N$ is not finitely presentable.
Best Answer
Yes.
(EDIT: If you don't like following links, this is the Wikipedia article on Free abelian groups which, uncharacteristically, contains a complete (and correct) proof of precisely that statement).