z' = z'(x); z is function of x
c= constant.
Function is F = z'{^n} e^y $
and this the operation needed to be done.
$ z' (\frac{dF}{dz'}) - F = c $
so I started with.
$ = [z' n ( z'^{n-1} e^y)] -[z'{^n} e^y] = c $
z' from outsdide in first term multiplies to z' inside the small brackets
$ [ n(z'^{n-1+1} e^y) ] - [z'^n e^y] = c$
$ [ n(z'^{n} e^y] - [z'^n e^y] = c$
(Q1- is that right so far ?) doing minus operation gives
$ (n-1)(z'^n e^y) = c $
now to get z out of z'
$ \int dx (n-1)(z')^n = \int dx \frac{c}{e^y}$
(Q2: what happens next ?)
how do I get? $ z=n \, ln((constant)\, x+e^{1/n} ? $
BC is $ y(0) =1 , y(1)= A $
where $ constant = e^{A/n} - e^{1/n} $
First, explaining the meaning/origin of your equation can be a good idea.
Next: all OK until $$(n-1)(z'^n e^y) = c.$$ But now... $y$ is $y$ (another variable $\ne x$) or is actually $x$? In any case, $$z' = \root n\of{\frac{ce^{-y}}{(n-1)}} = \root n\of{\frac{c}{(n-1)}}e^{-y/n}.$$ And if $y$ is actually $x$, integrating: $$z = -n\root n\of{\frac{c}{(n-1)}}e^{-x/n}+k.$$