Question 15349
 You are right for the inverse of
 f(x)= y = {{{(x^3+8)/3}}}
 to get 
 y = {{{ f^(-1)(x) = (3x-8)^(1/3) }}}
 
 But use ^ for exponent and parentehsis for grouping terms next time. 

 If the original function is constructed by elementary invertible
 function (such as addition, multiplication,...) , we only need to
 apply reverse process step by step to get the inverse function.
        
  Since y=f(x)= {{{T[k]}}}{{{T[k-1]}}}..{{{T[1](x)}}} then
  {{{f^(-1)(x)}}} = {{{T[1]^(-1)}}} {{{T[2]^(-1)}}}..{{{T[k]^(-1)(x)}}}
 (if each {{{T[i] }}} is invertible.]

 By for some functions such as y = x/(x^2+1) (or the functions containing
 some transcendental functions as trig or log ) 
 is not so easy to get 1st inverse. I suggest that you try.
  
 Anyway, you have got good idea and good luck!!!

 Kenny