Find the solution:

$u_{x}+3u_{y}=xy$

$u_{x=0}=0$

By examining Integral Lines:

$\frac{1}{dx}=\frac{3}{dy}=\frac{xy}{du}$ <very ugly form. Never use it $\color{blue} {\frac{dx}{1}=\frac{dy}{3}=\frac{du}{xy}}$

Then from the first equality:

$3x=y+C$ where C is some constant.

$y=3x-C$

Then again from the Integral Lines:

$dx(xy)=du$

$dx(x(3x-C))=du$

$u=x^{3}-\frac{C}{2}x^{2}+C_{1}$

Then by the initial condition:

$u(x=0)=0$

$C_{1}=0$

Therefore,$C=3x-y$

Therefore,

$u(x,y)=x^{3}-\frac{1}{2}x^{2}C$

$u(x,y)=x^{3}-\frac{1}{2}x^{2}(3x-y)$

$u(x,y)=-\frac{1}{2}x^{3}+\frac{1}{2}x^{2}y$

Check:

$u_{x}(x,y)=-\frac{3}{2}x^{2}+xy$

$u_{y}(x,y)=\frac{1}{2}x^{2}$

$u_{x}+3u_{y}=xy$