Question 1025281
If y = f(x), then {{{df^(-1)(y)/dy = 1/(dy/dx) = 1/(df(0)/dx)}}}.  (A statement of the inverse function theorem. f(x) is continuously differentiable at x = 0.)

Now {{{f(x) = 2x + e^x}}} ==> f'(x) = {{{df(x)/dx = 2 + e^x }}}
==> {{{df^(-1)(1)/dy = dg(1)/dy = 1/(df(0)/dx) = 1/(2 + e^0) = 1/3}}}