I changed the format of your question, because I think you want to find the inflection points for #y= e^(2x) - e^x #

(Not for #y= e^2x - e^x # which has none.)

#y= e^(2x) - e^x#

#y'= 2e^(2x) - e^x#

#y'' = 4e^(2x) - e^x# . The partition numbers for #y''# are found by:

#4e^(2x) - e^x = e^x(4e^x-1) = 0# .

Since #e^x# is never #0#, the only partition number is the solution to:

#(4e^x-1) = 0#

#e^x = 1/4#

#x=ln(1/4) = -ln4 = -2ln2#

Return to consider the sign of

#y'' = e^x(4e^x - 1)# .

The first factor is always positive and the second is negative for #x < ln(1/4)# and positive for #x > ln(1/4)#, so the sign of #y''# does change at the partition number.

The only inflection point is at the point where #x = ln(1/4)#

At #e^x = 1/4#, it is easy to calculate #y# because #e^(2x) = (1/4)^2 = 1/16#, so

#y =1/16 - 1/4 = -3/16#

(Or, if you prefer, rewrite #y = e^x(e^x+1)# and use that to find #y#.)

The only inflection point is #(-ln4, -3/16)#