Step 1:
To check whether the given problem is a homogenous function of degree zero
we rewrite the given equation as $\large\frac{dx}{dy} = -\large\frac{ e^{\Large\frac{x}{y}}(1 - \large\frac{x}{y})}{1 + e^{\Large\frac{x}{y}}}$
Let $F(x,y) = -\large\frac{ e^{\Large\frac{x}{y}}(1 - \large\frac{x}{y})}{ 1+ e^{\Large\frac{x}{y}}}$
$F(kx,ky) =\large\frac{ e^{\Large\frac{-kx}{ky}}(1-\large\frac{kx}{ky})}{1 + e^{\Large kx/ky} }= -\large\frac{ e^x/y(1 - x/y)}{1 + e^x/y}$$= k^0.F(x,y) $
Hence this is a homogenous differential equation.
Step 2:
To solve this put $x = vy$ and $\large\frac{dx}{dy}$$ = v + y\large\frac{dv}{dy}$
So $v + y\large\frac{dv}{dy }= -e^{\Large\frac{vy}{y}}$$(1- vy/y)/1+e^{\Large\frac{ vy}{y}}$
Cancelling the common factor y we get
$v + v \large\frac{dv}{dy} = - e^v\large\frac{(1 - v)}{1+e^v}$
$v\large\frac{dv}{dy}= -e^v\large\frac{(1 - v)}{1+e^v}$$ -v$
$v\large\frac{dv}{dy} = v+\large\frac{ e^v}{1+e^v}$
Step 3:
Bringing v and dv to the LHS and x and dx to the RHS we get,
$[\large\frac{1+e^v}{v+e^v} ] = \frac{- dy}{y}$
Integrating on both sides we get,
$\log(v+e^y) = - \log y + \log C$
Step 4:
Substituting back for v we get
$\large\frac{x}{y} + e^{\Large\frac{x}{y }}= \frac{C}{y}$
On cancelling y throughout,we get the solution as
$x + ye^{\Large\frac{x}{y}} $$= C$