(1+e^x/y)dx + e^x/y ( 1-x/y )dy =0
$\displaystyle \displaystyle{\left(1 + e^{\frac{x}{y}}\right)\,dx + e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)\,dy = 0}$
If this is an exact equation of the form $\displaystyle M\,dx + N\,dy = 0$ then $\displaystyle \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.
$\displaystyle \frac{\partial M}{\partial y} = -\frac{x}{y^2}e^{\frac{x}{y}}$.
$\displaystyle \frac{\partial N}{\partial x} = -\frac{1}{y}e^{\frac{x}{y}} + \frac{1}{y}e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)$
$\displaystyle = -\frac{1}{y}e^{\frac{x}{y}} + \frac{1}{y}e^{\frac{x}{y}} - \frac{x}{y^2}e^{\frac{x}{y}}$
$\displaystyle = -\frac{x}{y^2}e^{\frac{x}{y}}$
$\displaystyle = \frac{\partial M}{\partial y}$ as required.
So it's an exact equation.
That means that the solution $\displaystyle F(x, y)$ has the property that
$\displaystyle \frac{\partial F}{\partial x} = M$ and $\displaystyle \frac{\partial F}{\partial y} = N$.
Thus $\displaystyle \frac{\partial F}{\partial x} = 1 + e^{\frac{x}{y}}$
$\displaystyle F = \int{1 + e^{\frac{x}{y}}\,dx}$
$\displaystyle = x + y\,e^{\frac{x}{y}} + g(y)$.
$\displaystyle \frac{\partial F}{\partial y} = e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)$
$\displaystyle F = \int{e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)\,dy}$
$\displaystyle = y\,e^{\frac{x}{y}} + h(x)$ (according to Wolfram)...
Comparing these two results, it's easy to see that $\displaystyle h(x) = x + C$ and $\displaystyle g(y) = C$.
Therefore $\displaystyle F =x + y\,e^{\frac{x}{y}} + C$.