[最も好ましい] (x^2-y^2-z^2)p 2xyq=2xz 136087-(x^2-y^2-z^2)p+2xyq=2xz

 I was solving 1st order PDE which was \begin{equation} (x^2y^2z^2)p2xyq=2xz \tag{1} \end{equation} I had tried to solve this Please tell me whether it is correct or not $$\frac{dx}{x^2y^2Find also the magnitude of this maximumA) \( \Large \phi \left(xyz, \frac{y}{z}\right)=0\) B) \( \Large \phi \left(\frac{y}{z},\frac{y}{x^{2}y^{2}z^{2}}\right) =0\) C) \( \Large \phi \left(\frac{y}{2

1

(x^2-y^2-z^2)p+2xyq=2xz