Answer
$ \frac{2}{3d-1}, with$ $d\ne\frac{1}{3}$
Work Step by Step
Given : $\frac{\frac{3}{3d^{2}+5d-2}}{\frac{3}{2d+4}}$
This becomes : $\frac{3}{3d^{2}+5d-2} \times \frac{2d+4}{3}$
$= \frac{2(d+2)}{(3d-1)(d+2)}$
$= \frac{2}{3d-1}$
(After dividing out common factor (d+2))
