Answer
$\dfrac{\partial P}{\partial V}\dfrac{\partial V}{\partial T}\dfrac{\partial T}{\partial P}=-1$
Work Step by Step
Since, $P= \dfrac{mRT}{V} ,V=\dfrac{mRT}{P}; T= \dfrac{PV}{mR}$ ...(1)
$\dfrac{\partial P}{\partial T}=\dfrac{\partial}{\partial T}[\dfrac{mRT}{V}]=\dfrac{mR}{V}$ ...(2)
and $\dfrac{\partial T}{\partial P}=\dfrac{V}{mR}$ ...(3)
From equations (1), (2) and (3), we have
$\dfrac{\partial P}{\partial V}\dfrac{\partial V}{\partial T}\dfrac{\partial T}{\partial P}=(\dfrac{-mRT}{V^{2}}) (\dfrac{mR}{P})(\dfrac{V}{mR})$
$\implies \dfrac{-mRT}{PV}=\dfrac{-mRT}{\dfrac{mRT}{V}\times V}=-1$
Thus, $\dfrac{\partial P}{\partial V}\dfrac{\partial V}{\partial T}\dfrac{\partial T}{\partial P}=-1$
