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
$$\frac{xdx-ydy}{2x^2-2y^2}=\frac{xdx+ydy}{4x^2+4y^2} \\ \frac{d(x^2-y^2)}{4(x^2-y^2)}=\frac{d(x^2+y^2)}{8(x^2+y^2)} \\ \enclose{box}{\frac{(x^2-y^2)^2}{x^2+y^2}=C_2}$$