Let $p\equiv 1\,(\mathrm{mod}\,9)$ be a prime number and $\zeta_3$ primitive cube root of unity. Then $\mathrm{k}=\mathbb{Q}(\sqrt[3]{p},\zeta_3)$ is pure metacyclic field with group $\mathrm{Gal}(\mathrm{k}/\mathbb{Q})\simeq S_3$. In the case that $\mathrm{k}$ possesses $3$-class $C_{\mathrm{k},3}$ type $(9,3)$, capitulation $3$-ideal classes in its unramified cyclic cubic extensions determine...