Since the domain of #f# is #RR-{3}#, #x# cannot take the value of #3#.
The point where #nx-9=0# is not in the domain, therefore #3n-9=0# is not in the domain.
#3n-9=0#
#<=>#
#3n=9#
#<=>#
#n=3#
So, by this point we have #f(x)=(mx+4)/(3x-9)#
If #f# is one-to-one and onto, then #f# is bijective. If #f# is bijective, then it has an inverse.
#EE f^(-1):RR-{4} to RR-{3}#
First we need to find it
Let #y=f(x)# then
#y=(mx+4)/(3x-9)#
#<=># multiply both sides by #3x-9#
#(3x-9)y=(mx+4)#
#<=># distribute #y#
#3xy-9y=mx+4#
#<=># subtract #4# from both sides
#3xy-9y-4=mx#
#<=># subtract #mx# from both sides
#3xy-9y-4-mx=0#
#<=># group like terms
#3xy-mx-9y-4=0#
#<=># factor #x# from #3xy-mx#
#x(3y-m)-9y-4=0#
#<=># subtract #x(3y-m)# from both sides
#-9y-4=-x(3y-m)#
#<=># divide both sides by #(3y-m)#
#-(9y-4)/(3y-m)=-x#
#<=># multiply both sides by #-1#
#(9y-4)/(3y-m)=x #
and since #(f^(-1)@f)(x)=f^(-1)(f(x))=x#
#=> f^(-1)(x)=(9x-4)/(3x-m)#
and since the domain of #f^(-1)# is #RR-{4}#, we have that
#12-m=0#
#<=>#
#12=m#
Then this leads us to #f(x)=(12x+4)/(3x-9)#
Then #f(1/3)=(12(1/3)+4)/(3(1/3)-9)=(4+4)/(1-9)=8/(-8)=-1#
