y=(ex-e-x)/(ex+e-x)=tanh(x), dy/dx=sech2(x)=1-tanh2(x)=1-y2.
Without using hyperbolic functions we can still prove the DE.
y2=(e2x-2+e-2x)/(e2x+2+e-2x), 1-y2=4/(e2x+2+e-2x)=(2/(ex+e-x))2.
Let u=ex-e-x, v=ex+e-x then du/dx=ex+e-x, dv/dx=ex-e-x,
y=u/v, dy/dx=(vdu/dx-udv/dx)/v2.
dy/dx=[(ex+e-x)2-(ex-e-x)2]/(ex+e-x)2=(2/(ex+e-x))2=1-y2 QED