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)))