Find general solution to this PDE $$x^2z{\partial z\over\partial x} + y^2z{\partial z \over \partial y} = x+y$$ 
My attempt: 
$${dx\over x^2z}={dy\over y^2z}={dz\over x+y}$$
Notice ${dx\over x^2z}={dy\over y^2z} \text{ we can multiply by } z \text{ and after integration} \Rightarrow {1\over x} = {1\over y} + C_1$
Here I stuck, 'cause I can't find new combination to integrate. I'm sure there are a lot of people quite good at finding these combination. So my question is also what are you looking at first. Is there common tricks to get this done? 

The beginning of your calculus is correct. Next step :
$${dx\over x^2}={dy\over y^2}=\frac{dx-dy}{x^2-y^2}\qquad\text{see note below}.$$
$${dx\over x^2z}={dy\over y^2z}=\frac{dx-dy}{(x+y)(x-y)z}={dz\over x+y}$$
$$\frac{dx-dy}{(x-y)}=zdz$$
$$ z^2-2\ln|x-y|=C_2$$
General solution :
$$z^2-2\ln|x-y|=F\left(\frac{1}{x}-\frac{1}{y} \right)$$
$F$ is an arbitrary function, to be determined according to some boundary condition.
$$z(x,y)=\pm\sqrt{2\ln|x-y|+F\left(\frac{1}{x}-\frac{1}{y} \right)}$$
NOTE :
Well-known property of the fractions :
$\frac{A}{B}=\frac{C}{D}=\frac{k_1A+k_2C}{k_1B+k_2D}$
In the above equation : $k_1=1$ and $k_2=-1$ , chosen in order to simplify further calculus.
