As #x^2# ranges from #[0,oo)#,
#2-x^2# ranges from #(-oo,2)#.
Observe that we can not have #x^2=2# or #x=+-sqrt2# as that makes #2-x^2=0#, which appears in denminator in #f(x)# and we have #x=sqrt2# and #x=-sqrt2# as asymptotes
Now for
(a) #x in(-oo,-sqrt2)#, #2/(2-x^2)# ranges from #(-oo,0)#
(b) #x in(-sqrt2,sqrt2)#, #2/(2-x^2)# ranges from #(1,oo)# and
(c) #x in(sqrt2,oo)#, #2/(2-x^2)# again ranges from #(-oo,0)#
Hence range of #f(x)=2/(2-x^2)# is #RR# except #[0,1)#
graph{2/(2-x^2) [-10, 10, -5, 5]}
