Solving $x^2u_x + y^2u_v + z(x+y)u_z = 0$ Exercise :

Solve the PDE
  $$x^2u_x + y^2u_v + z(x+y)u_z = 0$$

Attempt :
We have to solve the problem :
$$\frac{dx}{x^2} = \frac{dy}{y^2} = \frac{dz}{z(x+y)}$$
Choosing the first two fractions, we yield :
$$\frac{dx}{x^2} = \frac{dy}{y^2} \Rightarrow \int \frac{dx}{x^2} = \int \frac{dy}{y^2} \Leftrightarrow -\frac{1}{x} = \frac{1}{y} + c $$
$$\implies $$
$$u_1 = \frac{1}{x} - \frac{1}{y}$$
Now, choosing the first and the last fraction, we yield : 
$$\frac{dx}{x^2} = \frac{dz}{z(x+y)} \Rightarrow \int z(x+y)dx = \int x^2dz \Leftrightarrow z\frac{x^2}{2} + zxy = x^2z + c'  $$
but this is a different result for $u_2$ and eventually the general solution $u(x,y,z) = F(u_1,u_2)$ than Wolfram Alpha calculates here. Why is this the case and where am I wrong with my solution ?

For the first equation
$$\frac{dx}{x^2} = \frac{dy}{y^2} \Rightarrow \int \frac{dx}{x^2} = \int \frac{dy}{y^2} \Leftrightarrow -\frac{1}{x} = -\frac{1}{y} + C$$
$$ \implies  K=\frac 1x- \frac 1y $$
For the last equation use this trick @Rebellos
$$\frac{dx}{x^2} = \frac{dy}{y^2} = \frac{dz}{z(x+y)}$$
$$\frac{dx-dy}{x^2-y^2} = \frac{dz}{z(x+y)}$$
$$\frac{d(x-y)}{x-y} = \frac{dz}{z}$$
I let you finish this last equation
