First order linear differential equation (possible error in book) Consider following differential equation in $$\frac{dx}{(3x^2+y^2)/x}=\frac{dy}{(3y^2+x^2)/y}=\frac{dz}{-(x^2+y^2)/z}$$
$\require{enclose}$
Taking multipliers $x, y, 4z$ we get $$xdx+ydy+4zdz=0 \\ \enclose{box}{x^2+y^2+4z^2=C_1}$$
I have this function correct. 
Book writes: for taking first two fraction $xdx-ydy=0$ and hence $$\bbox[5px,border:2px solid red]{x^2-y^2=C_2}$$
I am sure this is the correct solution. 
We would have on subtracting numerator and denominator $$\frac{xdx-ydy}{2x^2-2y^2}$$
for which denominator isn't 0 so how can we assert $xdy-ydx=0$
I am not missing anything am I?
Possible Solution 
$$\frac{xdx-ydy}{2x^2-2y^2}=\frac{xdx+ydy}{4x^2+4y^2} \\ \frac{d(x^2-y^2)}{4(x^2-y^2)}=\frac{d(x^2+y^2)}{8(x^2+y^2)} \\ \enclose{box}{\frac{(x^2-y^2)^2}{x^2+y^2}=C_2}$$

You are right, two charecteristic equations are :
$$x^2+y^2+4z^2=C_1$$
$$\frac{(x^2-y^2)^2}{x^2+y^2}=C_2$$
From that, the general solution of the PDE expressed on the form of implicit equation $C_1=F(C_2)$ with arbitrary function $F$ :
$$x^2+y^2+4z^2=F\left(\frac{(x^2-y^2)^2}{x^2+y^2}\right)$$
On explicit form :
$$z(x,y)=\pm\frac12\sqrt{-x^2-y^2+F\left(\frac{(x^2-y^2)^2}{x^2+y^2}\right)}$$
NOTE : 
$xdx-ydy=0$ is not correct in the general case thus $x^2-y^2=C_3$ is not a correct characteristic equation. 
NOTE :
The problem is simplified with the change of variables 
$\quad\begin{cases} X=x^2\\ Y=y^2\\ Z=z^2 \end{cases}$ :
$$\frac{dX}{3X+Y}=\frac{dY}{3Y+X}=-\frac{dZ}{X+Y}$$
