It should be easy to see that e^{x} and e^{-x} are two independent solutions to y"- y= 0.
The "variation of parameters" is to look for a solution of the form
y(x)= u(x)e^{x}+ v(x)e^{-x}
In fact, there are an infinite number of u(x), v(x) that would work.
Plug that into the original equation and, because e^{x} and e^{-x} satisfy the original homogeneous equation, the "u(x)" and "v(x)" terms cancel leaving just u'e^{x}- v'e^{-x}= 2/(1+e^{x}). Treat that, along with
u'e^{x}+ v'e^{-x}= 0 (above) as two linear equations for u', v'.
Integrate those solutions to find u and v and plug into y(x)= u(x)e^{x}+ v'(x)e^{-x}.