= 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 $a,d\ne0$, 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 $ad-bc$, proving
$$
\boxed{\left(\frac{\partial S}{\partial T}\right)_p\left(\frac{\partial p}{\partial V}\right)_T=\left(\frac{\partial p}{\partial V}\right)_S\left(\frac{\partial S}{\partial T}\right)_V.}
$$
The <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.
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