trying to get the second derivative from this
$\displaystyle f'(x)=-[{e^{x}(x-2)}/{(x-1)^{2}}]$
been using the quotient rule but answer differs from book
Alright, using the quotient rule you should get that:
$\displaystyle f''(x) = -\frac{2(x - 1)e^x(x - 2) - [e^x(x - 2) + e^x](x - 1)^2}{(x - 1)^4}$
Note: In the second term in the numerator you have to separately use the product rule or this will not come out right.
$\displaystyle f''(x) = -\frac{2(x -1)e^x(x - 2) - e^x(x - 1)^3}{(x - 1)^4}$
From here you can simplify it however you like, but here is an option:
$\displaystyle f''(x) = -\frac{e^x[2(x - 2) - (x - 1)^2]}{(x - 1)^3} = -\frac{e^x(x+5)}{(x - 1)^2}$