PDE : $x^2 z_x + y^2 z_y = z(x+y)$

From

$$ \frac{1}{x}-\frac{1}{y}=C_1 \Rightarrow y-x=C_1 x y $$

From

$$ \frac{dy-dx}{y-x}=\frac{dz}{z}\Rightarrow y-x = C_2 z $$

hence

$$ C_2 z = C_1 x y \Rightarrow C_4 = \frac{z}{x y} $$