Prove that $\frac{\partial x}{\partial y} \frac{\partial y}{\partial z} \frac{\partial z}{\partial x} = -1$ and verify ideal gas law Ok guys, continuing my passage through edwards... here is the question... thanks for hints/solutions in advance:
Suppose $f(x,y,z)=0$ can be solved for each of the three variables $x,y,z$ as a differentiable function of the other two. Then prove that
$\displaystyle \frac{\partial x}{\partial y} \frac{\partial y}{\partial z} \frac{\partial z}{\partial x} = -1$
Verify this is the case for the ideal gas equation $pv =RT$ where (where $p,v,T$ are the three variables and $R$ is the constant).
 A: Define $f(x(y,z),y,z)=f(x,y(x,z),z)=f(x,y,z(x,y))=0$. Then by the
chain rule, 
\begin{eqnarray*}
\frac{\partial f}{\partial y}+\frac{\partial f}{\partial x}\frac{\partial x}{\partial y} & = & 0\\
\frac{\partial f}{\partial z}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial z} & = & 0\\
\frac{\partial f}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial x} & = & 0
\end{eqnarray*}
So that 
\begin{eqnarray*}
\frac{\partial x}{\partial y} & = & -\frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}\\
\frac{\partial y}{\partial z} & = & -\frac{\frac{\partial f}{\partial z}}{\frac{\partial f}{\partial y}}\\
\frac{\partial z}{\partial x} & = & -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}
\end{eqnarray*}
Hence, 
$$
\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x}=-\frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}\frac{\frac{\partial f}{\partial z}}{\frac{\partial f}{\partial y}}\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}=-1.
$$
