Solve the differential equation $y{e^{\frac{x}{y}}}dx = \left( {x{e^{\frac{x}{y}}} + {y^2}} \right)dy(y \ne 0)$

$y{e^{\frac{x}{y}}}dx = \left( {x{e^{\frac{x}{y}}} + {y^2}} \right)dy$
$\Rightarrow$ $y{e^{\frac{x}{y}}}\frac{{dx}}{{dy}} = x{e^{\frac{x}{y}}} + {y^2}$

$\Rightarrow$ ${e^{\frac{x}{y}}}\left[ {y \cdot \frac{{dx}}{{dy}} - x} \right] = {y^2}$
$\Rightarrow$ ${e^{\frac{x}{y}}} \cdot \frac{{\left[ {y \cdot \frac{{dx}}{{dy}} - x} \right]}}{{{y^2}}} = 1$

……(1)
Let ${e^{\frac{x}{y}}} = z$
Differentiating it with respect to ${\rm{y}}$,

we get:
$\frac{d}{{dy}}\left( {{e^{\frac{x}{y}}}} \right) = \frac{{dz}}{{dy}}$
$\Rightarrow$ ${e^{\frac{x}{y}}} \cdot \frac{d}{{dy}}\left( {\frac{x}{y}} \right) = \frac{{dz}}{{dy}}$

$\Rightarrow$ ${e^{\frac{x}{y}}} \cdot \left[ {\frac{{y \cdot \frac{{dx}}{{dy}} - x}}{{{y^2}}}} \right] = \frac{{dz}}{{dy}}$
From equation (1) and equation (2),

we get:
$\frac{{dz}}{{dy}} = 1$
$\Rightarrow$ $dz = dy$
Integrating both sides,

we get:
$z = y + C$
$\Rightarrow$ ${e^{\frac{x}{y}}} = y + C$