Use the chain rule.
#u(x) = e^x + (1+e^(2x))^(1/2) and y = ln(u)#
#(dy)/(du) = 1/u = 1/(e^x + (1+e^(2x))^(1/2))#
#(du)/(dx) = e^x + d/(dx)((1+e^(2x))^(1/2))#
For the square root use chain rule again with
#phi = (1+e^(2x))^(1/2)#
#v(x) = 1 + e^(2x) and phi = v^(1/2)#
#(dv)/(dx) = 2e^(2x) and (dphi)/(dv) = 1/(2sqrt(v))#
#(dphi)/(dx) = (dphi)/(dv)(dv)/(dx) = (e^(2x))/(sqrt(1 + e^(2x)))#
#therefore (du)/(dx) = e^x + (e^(2x))/(sqrt(1 + e^(2x)))#
#(dy)/(dx) = (dy)/(du)(du)/(dx)#
#=1/(e^x + (1+e^(2x))^(1/2))*(e^x + (e^(2x))/(sqrt(1 + e^(2x))))#
#=e^x/(e^x + sqrt(1+e^(2x))) + e^(2x)/(sqrt(1 + e^(2x))(e^x+sqrt(1+e^(2x)))#
Bringing together over LCD:
#=(e^xsqrt(1+e^(2x)) + e^(2x))/(sqrt(1 + e^(2x))(e^x+sqrt(1+e^(2x)))#
Take factor of #e^x# out of numerator:
#=(e^x(sqrt(1+e^(2x)) + e^x))/(sqrt(1 + e^(2x))(e^x+sqrt(1+e^(2x)))#
Cancel out and obtain
#=(e^x)/(sqrt(1+e^(2x)))#