Question 192313
Clearly,The function is  not defined at the pre image 3x-5=0 i.e x=5/3, because the remainder is zero at this value and the value of the function become meaningless. If X does not equal 5/3 then the function is defined in the whole of Real numbers.Therefore the domain is R-{5/3}
Let y=2x+1/3x-5 
then 3xy-5y=2x+1,
then x(3y-2)=5y+1,
then x=5y+1/3y-2
From this relation it is clear that x is not defined when y=2/3 (i.e 3y-2=0) Therefore the Range of the function is R-{2/3}
Notation:R is the set of all real numbers and R-{2/3} is the set of all real numbers excluding 2/3