Question

dx/(x^2-y z) =dy/(y^2-zx)=dz/(z^2-xy)

Answer 1

$$\dfrac{dx }{x^2-yz}=\dfrac{dy}{y^2-zx}=\dfrac{dz}{z^2-xy}$$

Taking the multipliers $y, z, x,$ then the ratio,

$$\dfrac{ydx+zdy+xdz }{x^2y-y^2z+y^2z-z^2x+xz^2-x^2y}$$

$$=\dfrac{ydx+zdy+xdz }{0}$$

Hence

$$ydx+zdy+xdz=0$$

Integrate

$$xy+yz+zx=c_1$$

Taking the multipliers $xy, xz, x^2,$ then the ratio,

$$\dfrac{xydx+xzdy+x^2dz }{x^3y-xy^2z+xy^2z-z^2x^2+x^2z^2-x^3y}$$

$$=\dfrac{xydx+xzdy+x^2dz }{0}$$

Hence

$$xydx+xzdy+x^2dz=0$$

Integrate

$$\dfrac{x^2y}{2}+xyz+x^2z=\dfrac{c_2}{2}$$

$$x^2y+2xyz+2x^2z=c_2$$

Therefore the solutin is

$$\Phi(xy+yz+zx,x^2y+2xyz+2x^2z )=0$$

where $\Phi$ is an arbitrary function.