#dy/dx = -2/(e^x+e^-x)^2(e^x-e^-x )= -1/2y^2(e^x-e^-x)#
Differentiating once again, we get
#(d^2y)/dx^2 = -1/2 2y dy/dx(e^x-e^-x)-1/2y^2(e^x+e^-x)#
#qquad =-y(-1/2y^2(e^x-e^-x))(e^x-e^-x) -1/2y^2 2/y#
# qquad = 1/2y^3color(red)((e^x-e^-x)^2)-y#
Now
#(e^x+e^-x)^2-(e^x-e^-x)^2 = 4 implies#
#(2/y)^2-(e^x-e^-x)^2 = 4 implies#
#(e^x-e^-x)^2 = 4/y^2-4#
Thus
#(d^2y)/dx^2 = = 1/2y^3color(red)((4/y^2-4))-y = 2y-2y^3-y#
#qquad = y-2y^3#
Alternately,
Using hyperbolic functions can reduce some of the work here :
#y = sechx implies dy/dx = -sechx tanhx#
#(d^2y/dx^2) = (-sechxtanhx )tanhx-sechx(sech^2x)#
#qquad = -sechx(sech^2x-tanh^2x) #
#qquad = -sechx(sech^2x-(1-sech^2x))#
#qquad = -sech x(2sech^2x-1)#
#qquad = sechx-2sech^3x = y-2y^3#
