Lagrange's auxiliary equations are $\dfrac{dx}{xz}=\dfrac{dy}{-yz}=\dfrac{dz}{y^2-x^2}$ . The first pair gives $dx/x=-dy/y$ , integrating to $u=xy=$ const. Multiplying the first two by $x,y$ and adding, then combining with the third (multiplier $1$ on $dz$ ) gives $x\,dx+y\,dy+z\,dz=0$ , integrating to $v=x^2+y^2+z^2=$ const. So the true general solution is $\Phi(xy,\,x^2+y^2+z^2)=0$ for arbitrary $\Phi$ .

(A) is this in its most direct solved form, $v=f(u)$ . (D) is equivalent to it: writing $w=(x+y)^2+z^2=v+2u$ , the relation $w=f(u)$ says $v=f(u)-2u$ , itself an arbitrary function of $u$ — the same family as (A). (B) is also equivalent, using $(u,w)$ as an alternative (invertible) coordinate pair for the same two invariants in place of $(u,v)$ .

(C), however, introduces $y/x$ , which is not constant along the characteristics: for fixed $u=xy$ and $v=x^2+y^2+z^2$ , the ratio $y/x$ can still take different values (e.g. $(x,y)=(1,2)$ and $(x,y)=(2,1)$ give the same $xy=2$ but $y/x=2$ versus $y/x=0.5$ ). A relation built from $y/x$ is not respected by the actual characteristic curves of this PDE, so it does not give a general solution.