It should be easy to see that ex 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)ex+ 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 ex and e-x satisfy the original homogeneous equation, the "u(x)" and "v(x)" terms cancel leaving just u'ex- v'e-x= 2/(1+ex). Treat that, along with
u'ex+ 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)ex+ v'(x)e-x.