Step 1: Rewrite the given equation as:

$$\frac{dx}{dy} = \frac{-e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)}{1 + e^{\frac{x}{y}}}$$

Since the right-hand side is a function of $\frac{x}{y}$ alone, the equation is homogeneous.

Step 2: Substitute $x = vy$, so $\frac{dx}{dy} = v + y\frac{dv}{dy}$. This gives:

$$v + y\frac{dv}{dy} = \frac{-e^{v}(1-v)}{1+e^{v}}$$

Step 3: Simplify to get:

$$y\frac{dv}{dy} = \frac{-(e^{v}+v)}{1+e^{v}}$$

Separating variables:

$$\frac{1+e^{v}}{e^{v}+v}\,dv = -\frac{dy}{y}$$

Step 4: Integrate both sides: Let $u = e^{v}+v$, so $du = (1+e^{v})\,dv$. Thus:

$$\int\frac{du}{u} = -\int\frac{dy}{y}$$

giving:

$$\ln|e^{v}+v| = -\ln|y| + \ln C$$

Step 5: Simplify to get:

$$y(e^{v}+v) = C$$

Substituting back $v = \frac{x}{y}$:

$$y\left(e^{\frac{x}{y}} + \frac{x}{y}\right) = C$$

or:

$$x + ye^{\frac{x}{y}} = C$$

∴ The general solution is $x + ye^{\frac{x}{y}} = C$.