$$u_x+3u_y=xy\\
\frac{dx}{1} = \frac{dy}{3} = \frac{du}{xy}\\
3x-y = C\\
du=xydx=x(3x-C)dx=(3x^2-Cx)dx\\
u = x^3 - \frac{C}{2}x^2 + \phi(3x-y)\\
u = x^3 - \frac{3x-y}{2}x^2 + \phi(3x-y)\\
u|_{x=0}=\phi(-y)=0\\
u = x^3 - \frac{3x-y}{2}x^2=-\frac{1}{2}x^3+\frac{1}{2}x^2y
$$
