Answer
$y' = \frac{1 - e^y}{xe^y + 1}$
Work Step by Step
$xe^y = x - y$
use the product rule on $xe^y$ and differentiate both sides
$(x)(y'e^y) + (1)(e^y) = 1 - y'$
$xy'e^y + y' = 1 - e^y$
$y'(xe^y + 1) = 1 - e^y$
$y' = \frac{1 - e^y}{xe^y + 1}$