Calculus: Early Transcendentals, Chapter 3, 3.4, Section 3.4, Problem 28

Using the quotient rule of derivatives:
y' = [d/(du) (e^u-e^(-u)) * (e^u+e^(-u)) - (e^u-e^(-u)) d/(du) (e^u +e^(-u))]/(e^u+e^(-u))^2
= [(e^u -e^(-u) *(-1))*(e^u+e^(-u)) - (e^u-e^(-u))*(e^u+e^(-u) *(-1))]/(e^u+e^(-u))^2
=[(e^u + e^(-u))*(e^u+e^(-u))- (e^u-e^(-u))*(e^u-e^(-u))]/(e^u+e^(-u))^2
= [(e^u+e^(-u))^2-(e^u-e^(-u))^2] /(e^u+e^(-u))^2
= [(e^u +e^(-u)+e^u-e^(-u)) *(e^u+e^(-u)-e^u+e^(-u))]/(e^u+e^(-u))^2
= (4e^ue^(-u))/(e^u+e^(-u))^2
= 4/(e^u+e^(-u))^2

Hope this helps.