The DE is
$\frac {dx}{x^2-y^2-z^2}=\frac {dy}{2xy}=\frac {dz}{2xz}$
We can solve the 2nd and 3rd fraction to get $y=cz$.how would you get the other equation?
Yes, y= cz satisfies
I would prefer to write that as z= cy which also satisfies the equation. Then replacing z with cy in we have . Clearing the fractions, that is the same as or . That is not exact but it is not too difficult to find an integrating factor- since these are all powers of x and y, try an integrating factor of the form and use the exactness condition to determine that we must have n= 0, m= -2. That is, is an integrating factor. Multiplying by , we have . That is an exact equation because .