It is easy to check that for any value of $c$, the function $$y = ce^{-2x} + e^{-x}$$
is solution of equation $$y' + 2y = e^{-x}.$$ Find the value of $c$ for which the solution satisfies the initial condition $y(-5)= 6$
I start out by making it $y' =-2y+ e^{-x}$
this gives me $df/dy = -2y$ and $df/dx = e^{-x}$
I'm stuck here not sure what to do next.
You already have the formula $y = c e^{-2x} + e^{-x}$. All you need to do is find $c$ such that $y(-5) = 6$. So let's plug in $x = -5$ and the corresponding $y = 6$: $$ 6 = c e^{10} + e^{5} \iff c e^{-10} = 6 - e^{5} \iff c = e^{-10}(6 - e^{5}) \iff c = 6 e^{-10} - e^{-5}). $$ (The numerical value of $c$ is approximately $-0.000646555$.)
If you were to have the ODE $y' + 2y = e^{-x}$, then you would first solve to via the complimentary/auxiliary function and particular solution method to get $y = c e^{-2x} + e^{-x}$ for a constant $c$, and then plug in your initial condition to get $c$, as I've done above.