 Solution

At fixed entropy, the total differential in coordinates $(T,p)$ is $0=S_T{|}_{p}dT+S_p{|}_{T}dp$. Dividing by $dT$ along that path gives $S_T{|}_p=-S_p{|}_T{p}_T{|}_S$. The same argument in coordinates $(T,V)$ gives $S_T{|}_V=-S_V{|}_T{V}_T{|}_S$. These are the two required intermediate identities.

To prove the final result without cancelling a possibly vanishing thermal-expansion derivative, write $a=S_T{|}_V$, $b=S_V{|}_T$, $c=p_T{|}_V$, $d=p_V{|}_T$. In a regular thermodynamic region with $a,d\ne 0$, fixed pressure gives $V_T{|}_p=-c/d$, while fixed entropy gives $T_V{|}_S=-b/a$. The chain rule then gives $S_T{|}_p=a-bc/d$ and $p_V{|}_S=d-bc/a$. Both cross-products equal $ad-bc$, provingThe heat-capacity and compressibility ratio identity rewrites this as $C_p/C_V={\kappa }_T/{\kappa }_S$: keeping entropy fixed rather than temperature fixed changes the mechanical stiffness by the same factor as the two thermal responses.