Question

x(y^2+z)-y(x^2+z)q=(x^2-y^2)z, x+y=0, z=1

Answer 1

$\frac{dx}{x(y^2+z)}=\frac{dy}{-y(x^2+z)}=\frac{dz}{z(x^2-y^2)}$

$\frac{xdx+ydy-dz}{x^2y^2+x^2z-x^2y^2-zy^2-zx^2+zy^2}=\frac{xdx+ydy-dz}{0}$

$xdx+ydy-dz=0$

$\frac{x^2}{2}+\frac{y^2}{2}-z=c_1$

$\frac{dx/x+dy/y+dz/z}{y^2+z-x^2-z+x^2-y^2}=\frac{dx/x+dy/y+dz/z}{0}$

$lnx+lny+lnz=lnc_2$

$xyz=c_2$