How is $\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } $ equivalent to $\frac{ y dx + xdy - zdz}{0}=\frac{ xdx - ydy -zdz}{0}$? In the book of PDE by Kumar, it is given that 

However, I couldn't figure out how is 
$$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 }  $$
is equivalent to 
$$\frac{ y dx + xdy - zdz}{0 } = \frac{ x dx - y dy -z dz}{ 0}  .$$
 A: $$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 }  $$
Use the well known basic property of the fractions :
$$\text{if}\quad \frac{A}{B}=\frac{C}{D} \quad \text{then}\quad \frac{A}{B}=\frac{C}{D}=\frac{c_1A+c_2C}{c_1B+c_2D}$$
$c_1,c_2$are arbitrary constants (not both nul}.
This property is valid for more fractions :
$$\frac{A}{B}=\frac{C}{D}=\frac{E}{F}=\frac{c_1A+c_2C+c_3E}{c_1B+c_2D+c_3F}$$
In the case of Eq.$(1)$ with $c_1=y\quad;\quad c_2=x \quad;\quad c_3=-z$ :
$$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 }  =\frac{ydx+xdy-zdz}{yz(x+y)+xz(x-y)-z(x^2+y^2)}=\frac{ydx+xdy-zdz}{0}$$
This implies $ydx+xdy-zdz=0$.
On the same way, with $c_1=x\quad;\quad c_2=-y \quad;\quad c_3=-z$ :
$$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 }  =\frac{xdx-ydy-zdz}{xz(x+y)-yz(x-y)-z(x^2+y^2)}=\frac{xdx-ydy-zdz}{0}$$
This implies $xdx-ydy-zdz=0$.
A: Some notes about solving PDE
$$z(x+y)z_x+z(x-y)z_y=x^2+y^2\qquad (1)$$


*

*see
https://www.math24.net/first-integrals-page-2/
Example 4

*In our case two independent first integrals is:
$$y^2+2xy-y^2=C_1, \quad z^2-2xy=C_2$$

*General solution of PDE $(1)$ is
$$z^2=2xy+f(y^2+2xy-x^2)$$

*Let $u=z^2$. Then from $(1)$ we get linear PDE
$$(x+y)u_x+(x-y)u_y=2x^2+2y^2\qquad (2)$$
with general solution
$$u=2xy+f(y^2+2xy-x^2)$$

*$u_p=2xy$ is particular solution of $(2)$

*$y^2+2xy-y^2=C_1$ is general solution of ODE
$$\frac{dx }{x+y} = \frac{dy}{x-y}.$$
