I am new to floor functions and I can't see a way to find this exact inequality. I have studied the cases but I'm more interested in being able to find this particular inequality.
$$ -2 \leq 3 \left \lfloor 2x \right \rfloor -2\left \lfloor 3x \right \rfloor \leq 1$$
$2x-1<\lfloor 2x\rfloor\le2x\quad\color{blue}{(*)}$
$3x-1<\lfloor 3x\rfloor\le3x\quad\color{blue}{(**)}$
By multiplying $(*)$ by $3$ and by multiplying $(**)$ by $-2$, it follows that
$6x-3<3\lfloor 2x\rfloor\le6x$
$-6x\le-2\lfloor 3x\rfloor<-6x+2$
and, by adding the last two inequalities, we get that
$-3<3\lfloor 2x\rfloor -2\lfloor 3x\rfloor<2\;.$
Since $\;3\lfloor 2x\rfloor -2\lfloor 3x\rfloor\;$ is an integer, it follows that
$-2\le3\lfloor 2x\rfloor -2\lfloor 3x\rfloor\le1\;.$