The PDE is the following :
$$(xy)u_x+(2xy)u_y=(x+y)u$$
To be honest, I haven't quite understood the method and it seems to be used in different ways and that confuses me(in some solutions I see parametrizations going on). I managed to apply it in some problems using the following characteristics equations, which in this problem are written like this :
$$\frac{dx}{xy}=\frac{dy}{2xy}=\frac{du}{(x+y)u}$$
From the first equality we get:
$$\frac{dy}{dx}=2$$ $$=>y=2x+C_1=>C_1=y-2x$$
We can also get this:
$$\frac{du}{dx}=\frac{(x+y)u}{xy}=(1/y+1/x)u$$ Integrating I get something strange (from what I've seen at least).
$$ln|u|=x/y+ln|x|+C_2$$
Am on the right track? Can I use my initial equations to get a better relationship for C2? By cancelling (x+y) perhaps?
I should clarify that I'm also not an expert in the method, but I can tell you what went wrong.
Your first part up to $c_1 = y-2x$ is correct
For the second integral, you have $dy=2dx$, therefore
$$ \frac{du}{u} = \left(\frac{1}{y} + \frac{1}{x}\right)dx = \frac{1}{2y}dy + \frac{1}{x}dx $$
Integrating the above
$$ \ln |u| = \frac12 \ln |y| + \ln|x| + c_2 $$
Taking the exponential
$$ |u| = e^{c_2}|x|\sqrt{|y|} \implies u = \pm e^{c_2}x\sqrt{|y|} = Cx\sqrt{|y|} $$
The last step is to take $C = f(c_1)$ so that
$$ u = x\sqrt{y} f(y-2x) $$
You can usually drop the absolute value on $|y|$ for convenience, unless a boundary condition is given.