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

(N/A) Given the differential equation: $y e^{\frac{x}{y}} dx = (x e^{\frac{x}{y}} + y^2) dy$
Rearranging the terms,we get: $y e^{\frac{x}{y}} \frac{dx}{dy} = x e^{\frac{x}{y}} + y^2$
Subtracting $x e^{\frac{x}{y}}$ from both sides: $e^{\frac{x}{y}} (y \frac{dx}{dy} - x) = y^2$
Dividing by $y^2$: $e^{\frac{x}{y}} \frac{y \frac{dx}{dy} - x}{y^2} = 1$ --- $(1)$
Let $v = \frac{x}{y}$,then $e^v \frac{d}{dy}(\frac{x}{y}) = 1$.
Alternatively,let $z = e^{\frac{x}{y}}$.
Differentiating $z$ with respect to $y$: $\frac{dz}{dy} = e^{\frac{x}{y}} \cdot \frac{d}{dy}(\frac{x}{y}) = e^{\frac{x}{y}} \cdot \frac{y \frac{dx}{dy} - x}{y^2}$.
Substituting this into equation $(1)$,we get: $\frac{dz}{dy} = 1$.
Integrating both sides with respect to $y$: $\int dz = \int dy \Rightarrow z = y + C$.
Substituting back $z = e^{\frac{x}{y}}$,the general solution is: $e^{\frac{x}{y}} = y + C$.