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#