In a compound, atoms of element Y form hexagonal close-packed...

Let the number of atoms of element Y forming the hexagonal close-packed (hcp) lattice be $N$.

The number of octahedral voids in a close-packed structure is equal to the number of atoms forming the lattice, which is $N$.

Since atoms of element X occupy $\dfrac{2}{3}$ of the octahedral voids, the number of atoms of X is $\dfrac{2}{3}N$.

The ratio of the number of atoms of X to that of Y is $\dfrac{2}{3}N : N = 2 : 3$.

Thus, the formula of the compound is $\text{X}_2\text{Y}_3$.

Answer: $\text{X}_2\text{Y}_3$