Answer
$f'(x) = \frac{2x-1-(x-1)~ln(x-1)}{[1-ln(x-1)]^2~(x-1)}$
The domain is $~~(1, e+1)\cup (e+1, \infty)$
Work Step by Step
$f(x) = \frac{x}{1-ln(x-1)}$
We can differentiate $f(x)$:
$f'(x) = \frac{1-ln(x-1)-(x)(-\frac{1}{x-1})}{[1-ln(x-1)]^2}$
$f'(x) = \frac{1-ln(x-1)+\frac{x}{x-1}}{[1-ln(x-1)]^2}$
$f'(x) = \frac{\frac{(1-ln(x-1))(x-1)}{x-1}+\frac{x}{x-1}}{[1-ln(x-1)]^2}$
$f'(x) = \frac{x-x~ln(x-1)-1+ln~(x-1)+x}{[1-ln(x-1)]^2~(x-1)}$
$f'(x) = \frac{2x-1-(x-1)~ln(x-1)}{[1-ln(x-1)]^2~(x-1)}$
When we consider $ln(x-1)$, then it is required that $(x-1)\gt 0$
Then $x \gt 1$
Also:
$1-ln(x-1) \neq 0$
$ln(x-1) \neq 1$
$x-1 \neq e$
$x \neq e+1$
The domain is $~~(1, e+1)\cup (e+1, \infty)$
