For $a \in \mathbb{N}$ denote $u(a) = \left[ \frac{a}{2} \right]$, $v(a) = \left[ \frac{a}{3} \right]$. If $n = p_1^{a_1} \cdots p_k^{a_k}$ is the prime factorization of $n \in \mathbb{N}$, it is straightforward that $D_2(n) = (u(a_1)+1)\cdots(u(a_k)+1)$, $D_3(n) = (v(a_1)+1)\cdots(v(a_k)+1)$.

Define $a_1 = 2 \cdot 998$ and set $a_i = 2v(a_{i-1})$ for $i \ge 2$. (Only the first several terms of the infinite sequence $(a_i)$ will be used.) With this definition we have $u(a_i) = 998$ and $u(a_i) = v(a_{i-1})$ for $i \ge 2$.

Note also that $v(a_i) = \lfloor \frac{2}{3}v(a_{i-1}) \rfloor < v(a_{i-1})$ if $v(a_{i-1}) > 0$. Thus the sequence $(v(a_i))$ decreases, so its terms are $0$ for sufficiently large $i$.

Take the first index $k$ such that $v(a_k) = 0$ and define $n = p_1^{a_1} \cdots p_k^{a_k}$. Because $u(a_i) = v(a_{i-1})$ for $i \ge 2$,

$$D_2(n) = (998+1)(u(a_2)+1)\cdots(u(a_k)+1) = 999(v(a_1)+1)\cdots(v(a_{k-1})+1).$$

In addition $D_3(n) = (v(a_1)+1)\cdots(v(a_{k-1})+1)(v(a_k)+1)$ equals $(v(a_1)+1)\cdots(v(a_{k-1})+1)$ since the last factor $v(a_k)+1$ is $1$ due to $v(a_k)=0$. Therefore $D_2(n) = 999D_3(n)$.