Work with the powers of 3.

For the sake of contradiction, assume that such a partition is possible.

From the conditions, if $a,~b$ are in a set and $ab$, $\frac{a}{b}$ are in the interval $[3,~3^5]$, they are in the opposite set.

Let the sets be $A,~B$ and assume that 3 is in $A$.

So, $3\cdot3$ is in $B$.

So, $3^2\cdot3^2=3^4$ is in $A$.

$A$ has $3^4,~3$. So, $3^4\cdot3=3^5$ and $\frac{3^4}{3}=3^3$ are in $B$.

We have $3^2,~3^3,~3^5$ in $B$. But, $3^2\cdot3^3=3^5$. Contradiction.

So, there is no such partition.