Suppose that a particle in $\mathbb{R^3}$ stays on the surface $xy + yz + xz = 0$. If the particle is at the point $(2, −1, 2)$ and starts moving along the surface parallel to the $xz$-plane, how fast is $z$ changing with respect to $x$?
I think that the plane where the particle is is the plane $-x -z + xz = 0$ (it's the plane where $y = -1$). Then, i isolated $z$ and derivated $z$ in respect to $x$, and got $\frac{dz}{dx} = \frac{-1}{(x-1)^2}$, and i thought that this should be the answer.
But i'm not sure, do i have to compute this derivative? What should be the answer?
I'm also confused because this questions appears along with anothers chain rule questions, and i don't see how and why i would use chain rule here.
Thanks!