Let us find #y'#.
By repeatedly applying Chain Rule,
#y'=2[ln(1+e^x)]^1cdot[ln(1+e^x)]'#
#=2ln(1+e^x)cdot{1}/{1+e^x}cdot(1+e^x)'#
#=2ln(1+e^x)cdot{1}/{1+e^x}cdot e^x#
#={2e^xln(1+e^x)}/{1+e^x}#
Let us find #y''#.
By pulling out #2# and applying Quotient Rule,
#y''=2{[e^x ln(1+e^x)]'cdot(1+e^x)-e^x ln(1+e^x)cdot(1+e^x)'}/{(1+e^x)^2}#
by Product Rule,
#=2{[e^x cdot ln(1+e^x)+e^x cdot{e^x}/{1+e^x}] (1+e^x)-e^xln(1+e^x)cdot e^x}/{(1+e^x)^2}#
by factoring out #e^x# and simpliying the numerator,
#=2e^x{ln(1+e^x)+e^xln(1+e^x)+e^x-e^xln(1+e^x)}/{(1+e^x)^2}#
by cancelling out #e^xln(1+e^x)#'s,
#={2e^x[ln(1+e^x)+e^x]}/{(1+e^x)^2}#
I hope that this was helpful.
