Show that $ \frac{\partial x}{\partial y}\frac{\partial y}{\partial z} \frac{\partial z}{\partial x} = -1 $ 
If $f(x, y, z) = 0$ and $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} , \frac{\partial f}{\partial z}  \neq 0$, show that
$$ \frac{\partial x}{\partial y}\frac{\partial y}{\partial z} \frac{\partial z}{\partial x} = -1  $$

I understand that posting questions without showing your work is frowned upon. I just couldn't come up with anything.
 A: You question makes sense when $x=x(y, z)$, $y=y(x, z)$, $z=z(x, y)$ are functions which are implicitly defined by $F(x, y, z)=0$. We can indeed assure that's the case since the implicit function theorem tells us that if the partial derivatives are non-zero, as given, then $F(x, y, z)=0$ defines each variable as an implicit function of the others. Then, using the classic result from the same theorem:
$$
\frac{\partial x}{\partial y}=-\frac{F_y}{F_x}, \qquad
\frac{\partial y}{\partial z}=-\frac{F_z}{F_y}, \qquad
\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}
$$
We get:
$$
\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x}=
(-1)^3\frac{F_y}{F_x}\cdot\frac{F_z}{F_y}\cdot\frac{F_x}{F_z}=-1
$$
As desired.
A: $$\begin{array}{l}given:\left\{ \begin{array}{l}f\left( {x,y,z} \right) = 0\\z = z\left( {x,y} \right)\end{array} \right.\\\left\{ \begin{array}{l}dz = \frac{{\partial z}}{{\partial x}}dx + \frac{{\partial z}}{{\partial y}}dy\\\left\{ {dx = \frac{{\partial x}}{{\partial y}}dy + \frac{{\partial x}}{{\partial z}}dz} \right.\\ = \frac{{\partial z}}{{\partial x}}\left( {\frac{{\partial x}}{{\partial y}}dy + \frac{{\partial x}}{{\partial z}}dz} \right) + \frac{{\partial z}}{{\partial y}}dy\\ = \left( {\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial z}}} \right)dz + \left( {\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}} + \frac{{\partial z}}{{\partial y}}} \right)dy\\\left\{ {\left( {\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial z}}} \right) = 1} \right.\\ = dz + \left( {\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}} + \frac{{\partial z}}{{\partial y}}} \right)dy\end{array} \right.\\ \Downarrow \\\left( {\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}} + \frac{{\partial z}}{{\partial y}}} \right)dy = 0\\\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}} + \frac{{\partial z}}{{\partial y}} = 0\\\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}} =  - \frac{{\partial z}}{{\partial y}}\\\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}}\frac{{\partial y}}{{\partial z}} =  - 1\end{array}$$

This is a note from a physics teacher. (Not my work, I rearrange it.)
I dont guarantee this is correct, but I think it is.
