Deduction of cyclic relation in thermodynamics.

Question

I want know what theorem I need to prove \begin{equation*} \left(\frac{\partial \,x}{\partial \,y}\right)_{z}\left(\frac{\partial \,y}{\partial \,z}\right)_{x} = -\left(\frac{\partial \,x}{\partial \,z}\right)_{y} \end{equation*} implies \begin{equation*} \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. \end{equation*} What is the argument?

Answer 1

Use the inverse function theorem to show that $$ \left(\frac{\partial z}{\partial x}\right)_y = \left[\left(\frac{\partial x}{\partial z}\right)_y\right]^{-1}. $$