<s>[INST] simple 2 sides inequality 
$$2<\frac{x}{x-1}\leq 3$$

Is the only way is to multiple both sides by $(x-1)^2$?
so we get $2x^2-4x+2<x^2-x $ and $x^2-x<3x^2-6x+3$ which is $-x^2+3x-2$ and $-2x^2+5x-3<0$ so the sloutions are:
$1<x\leq \frac{3}{2}$ and  $1<x\leq 2$ so overall it is $1<x\leq\frac{3}{2}$
 [/INST] Multiplying $(x-1)^2\gt 0$ is a good idea, but you made mistakes after that.
$$2(x-1)^2\lt x(x-1)\iff x^2-3x+2\lt 0$$$$\iff (x-2)(x-1)\lt 0\iff 1\lt x\lt 2$$
and 
$$x(x-1)\le 3(x-1)^2\iff 2x^2-5x+3\ge 0$$$$\iff (x-1)(2x-3)\ge 0\iff x\le 1\ \text{or}\ x\ge\frac 32.$$
Hence, the answer is
$$\frac 32\le x\lt 2.$$
</s>