If this is true, there must be some cycle in the set $A$, such that if $b$ is any term in the cycle and $c$ is the next term, $c=2b$ or $c=\frac{b}{3}$. This is obviously impossible.

Caucasus MO 2020 Seniors P1 wrote: Determine if there exists a finite set $A$ of positive integers satisfying the following condition: for each $a\in{A}$ at least one of two numbers $2a$ and $\frac{a}{3}$ belongs to $A$. Consider the numbers with maximal $v_2$ in $A$ and take the least number between them , say $b$ , then $\frac{b}{3}\in{A}$ , which is absurd by the choice of $b$ .