#f(x)=2/(1-x^2)#

To find the Domain find where the function is undefined, i.e. #a/0#

To do this we set the denominator equal to zero and solve, notice it is a difference of squares:

#1-x^2=0#

#(1-x)(1+x)=0#

#x=1# and #x=-1# so the domain is all real numbers except those two:

#{x|x in RR, x!=1, x!=-1}#

Now the range, as #x# gets really big positive or negative the #1# becomes insignificant so we must only consider #1/-x#, sense we know as

#2/-x ->+-oo, f(x) -> 0# then:

#2/(1-x^2) ->+-oo, f(x) -> 0#

so we know for numbers above and below the asymptotes the range is #-oo " to " 0#, what about between the asymptotes? In that case the numbers input are #-1<=x<=1# so at 0 the range is 2 and the closer we get to -1 or 1 the range is #2/"some tiny number" -> oo#, so that range is #2 " to " oo#

Putting it all together:

#{f(x)|f(x) in RR, x<0 or x>=2}#

graph{2/(1-x^2) [-10, 10, -5, 5]}