\(f\circ g(x)=f\left(g(x)\right)=f\left(\dfrac{1}{x-1}\right)\) for \(x\ne1\) and \(f\left(\dfrac{1}{x-1}\right)=\dfrac{1}{\dfrac{1}{x-1}-2}=\dfrac{x-1}{3-2x}\) for \(3-2x\ne0\).
Thus, \(f\circ g(x)=\dfrac{x-1}{3-2x}\) for \((-\infty,1)\cup\left(1,\frac{3}{2}\right)\cup\left(\frac{3}{2},\infty\right)\).