option a) is correct.
$g\left(h\left(x\right)\right) =g\left(\frac{x}{x-1}\right)$
$=1-\frac{x}{x-1}$
$= \frac{-1}{x-1}$
$h\left(g\left(x\right)\right) =h(1-x)$
$=\frac{1-x}{-x}$
$\frac{g(h(x))}{h(g(x))} = \frac{x}{(1-x)(x-1)} = \frac{h(x)}{g(x)}$
option A)