The range of the function is all possible values of #f(x)# it can have. It can also be defined as the domain of #f^-1(x)#.
To find #f^-1(x)#:
#y=1/(x-1)^2#
Switch the variables:
#x=1/(y-1)^2#
Solve for #y#.
#1/x=(y-1)^2#
#y-1=sqrt(1/x)#
#y=sqrt(1/x)+1#
As #sqrt(x)# will be undefined when #x<0#, we can say that this function is undefined when #1/x<0#. But as #n/x#, where #n!=0#, can never equal zero, we cannot use this method. However, remember that, for any #n/x#, when #x=0# the function is undefined.
So the domain of #f^-1(x)# is #(-oo,0)uu(0,oo)#
It so follows that the range of #f(x)# is #(-oo,0)uu(0,oo)#.