Question

z(x+y)p+z(x-y)q=x²+y² where p=dz/dx
q=dz/dx (partial differentiation) (i couldn't find its symbol so used d only)

Answer 1

P=z(x+y)=zx+zy

Q=z(x-y)=zx-zy

R=x²+y²

$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}$

$\frac{dx}{zx+zy}=\frac{dy}{zx-zy}=\frac{dz}{x^2+y^2}$

Multiplyers: x,-y,-z

xdx-ydy-zdz=0

$\frac{x^2}{2}-\frac{y^2}{2}-\frac{z^2}{2}=C_1$

Multiplyers: -x,y,z

-xdx+ydy+zdz=0

$\frac{-x^2}{2}+\frac{y^2}{2}+\frac{z^2}{2}=C_2$