Partial Derivatives : Show that $\frac{∂x}{∂y}\frac{∂y}{∂z}\frac{∂z}{∂x}=-1$ Let $f : \mathbb{R}^3 \rightarrow \mathbb{R}$. How can i show that  

If $f(x,y,z)=0$ then $$\frac{∂x}{∂y}\frac{∂y}{∂z}\frac{∂z}{∂x}=-1 $$

Any help will be appreciated.
 A: To understand why $$
\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}
\frac{\partial z}{\partial x} = -1$$
One need to know what those symbols mean!
In above expression, $\frac{\partial x}{\partial y}$ is a short hand for $\frac{\partial X(y,z)}{\partial y}$ where $X(y,z)$ is the value of $x$ which solves the equation $f(x,y,z) = 0$ for given $y,z$. i.e., $X(y,z)$ is a function which satisfies $f(X(y,z),y,z) = 0$.
Partial differentiate against $y$ and apply chain rule, one get
$$\frac{\partial X(y,z)}{\partial y}\frac{\partial f}{\partial x}(x,y,z)|_{x=X(y,z)} + \frac{\partial f}{\partial y}(x,y,z)|_{x=X(y,z)} = 0
\quad\implies\quad
\frac{\partial x}{\partial y} \stackrel{def}{=} \frac{\partial X(y,z)}{\partial y} = -
\frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}$$
By a similar argument, we have
$$\frac{\partial y}{\partial z} =
-\frac{\frac{\partial f}{\partial z}}{\frac{\partial f}{\partial y}}
\quad\text{ and }\quad
\frac{\partial z}{\partial x} = -
\frac{\frac{\partial f}{\partial z}}{\frac{\partial f}{\partial z}}$$
Multiply these $3$ relations together, the identity follows:
$$\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}
\frac{\partial z}{\partial x}
= 
\left(-\frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}\right)
\left(-\frac{\frac{\partial f}{\partial z}}{\frac{\partial f}{\partial y}}\right)
\left(-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}\right)
= -1$$
