Another "Partial Derivatives in Thermodynamics" Question

[conservedcharge]

Hi all,

It seems I haven't completely grasped the use of Partial Derivatives in general; I have seen many discussions here dealing broadly with the same topic, but can't find the answer to my doubt. So, any help would be most welcome:

In Pathria's book (3rd ed.), equation (1.3.11) says:
P = \frac{\left( \frac{\partial S}{\partial V}\right )_{N,E} } {\left (\frac{ \partial S}{\partial E} \right)_{N,V}} = - \left( \frac{\partial E}{\partial V} \right)_{N,S}
My question is 2 fold:

1. How is he writing the first equality in the above equation?
2. What properties of partial derivatives are being used here to figure out the correct subscripts on the extreme right in the equation, given the subscripts in \frac{\left( \frac{\partial S}{\partial V}\right )_{N,E} } {\left (\frac{ \partial S}{\partial E} \right)_{N,V}}?

 

[Fightfish]

1. How is he writing the first equality in the above equation?
2. What properties of partial derivatives are being used here to figure out the correct subscripts on the extreme right in the equation, given the subscripts in \frac{\left( \frac{\partial S}{\partial V}\right )_{N,E} } {\left (\frac{ \partial S}{\partial E} \right)_{N,V}}?

He's using the triple product rule
\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

 

[Chestermiller]

The equality $P = - \left( \frac{\partial E}{\partial V} \right)_{N,S}$ comes from $dE = TdS-PdV+\mu dN$ by setting dS and dN equal to zero.

The equality $\frac{\left( \frac{\partial S}{\partial V}\right )_{N,E} } {\left (\frac{ \partial S}{\partial E} \right)_{N,V}} = - \left( \frac{\partial E}{\partial V} \right)_{N,S}$ comes from
$$dS=\left(\frac{\partial S}{\partial V}\right)_{E,N}dV+\left(\frac{\partial S}{\partial E}\right)_{V,N}dE+\left(\frac{\partial S}{\partial N}\right)_{E,V}dN$$ by setting dN and dS equal to zero.

 

[conservedcharge]

He's using the triple product rule
\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

Thanks @Fightfish , that helps.