Prove identity of partial derivatives I can not do the following problem:
Prove the identity:
$$\left( \frac{\partial x}{\partial y} \right)_{z}\left( \frac{\partial y}{\partial z} \right)_{x}\left( \frac{\partial z}{\partial x} \right)_{y}=-1$$
State the properties that must be $x=x(y,z)$, $y=y(x,z)$, $z=(x,y)$.
The truth is I do not know how to start, and they do not know how to interpret the functions $x$, $y$, $z$. Any help or explanation will be the most grateful.
 A: $x=x(y,z)$, $y=y(x,z)$, $z=(x,y)$
$$dx= (\frac{\partial x}{\partial y})_z dy + (\frac{\partial x}{\partial z})_y dz$$
$$dy= (\frac{\partial y}{\partial x})_z dx + (\frac{\partial y}{\partial z})_x dz$$
$$\therefore dx= (\frac{\partial x}{\partial y})_z [(\frac{\partial y}{\partial x})_z dx + (\frac{\partial y}{\partial z})_x dz] + (\frac{\partial x}{\partial z})_y dz$$
$$dx= (\frac{\partial x}{\partial y})_z (\frac{\partial y}{\partial x})_z dx + [(\frac{\partial x}{\partial y})_z(\frac{\partial y}{\partial z})_x + (\frac{\partial x}{\partial z})_y] dz$$
$$(\frac{\partial x}{\partial y})_z (\frac{\partial y}{\partial x})_z dx + [(\frac{\partial x}{\partial y})_z(\frac{\partial y}{\partial z})_x + (\frac{\partial x}{\partial z})_y] dz=1dx+0dz$$
$$
(\frac{\partial x}{\partial y})_z(\frac{\partial y}{\partial z})_x + (\frac{\partial x}{\partial z})_y=0
$$
Using reciprocal relation:
$$
(\frac{\partial x}{\partial y})_z = \frac{1}{(\frac{\partial y}{\partial x})_z}
$$
$$\left( \frac{\partial x}{\partial y} \right)_{z}\left( \frac{\partial y}{\partial z} \right)_{x}\left( \frac{\partial z}{\partial x} \right)_{y}=-1$$
A: The way I always thought about this was to pick one of the variables to be thought of as the dependent variable. Here I will pick $z$. Then we think of $z(x,y)$ to be a function which has partial derivatives $\partial_x z = \frac{\partial z}{\partial x}$ and $\partial_y z = \frac{\partial z}{\partial y}$. 
Now we must compute $$\left( \frac{\partial x}{\partial y} \right)_{z}\left( \frac{\partial y}{\partial z} \right)_{x}\left( \frac{\partial z}{\partial x} \right)_{y}.$$
Let's look at each term individually. The strategy will be to write each term in terms of the "regular" partial derivatives $\partial_x z$ and $\partial_y z$. These are "regular" in the sense that they are partial derivatives of the dependent variable with respect to the independent variable.
The third term is the easiest. It is just already a regular partial derivative $\partial_x z$. 
The second term is more foreign. If we are thinking about $z$ as a dependent variable, then it looks like we are taking the derivative of a independent variable with respect to a dependent variable. However, I just think of this as a shorthand for $1/\partial_y z$. 
The first term is the most complicated. Here it looks like we are taking the derivative of an independent variable with respect to another independent variable. You might think this would be zero, but $z$ is supposed to be held fixed. So the question is "If I change $y$ how much do I have to change $x$ to keep $z$ fixed?" Let's imagine we change $y$ by an amount $dy$. Then $z$ will change by an amount $\partial_y z dy$. To compensate we must change $x$ by an amount that will cause the opposite change ($-\partial_y z dy$). The correct $dx$ is given by the equation $\partial_x z dx = -\partial_y z dy$, so the amount we must change $x$ is $dx=\frac{-\partial_y z dy}{\partial_x z}$. Then $dx / dy =  -\partial_y z / \partial_x z$.
Now putting our three terms together we have $$\left( \frac{\partial x}{\partial y} \right)_{z}\left( \frac{\partial y}{\partial z} \right)_{x}\left( \frac{\partial z}{\partial x} \right)_{y}=-\frac{\partial_y z}{ \partial_x z}\frac{1}{\partial_y z} \partial_x z =-1. $$
