It's just the map 

$$x \mapsto y = \frac{x}{x^{\sigma}},$$

where the corresponding degree three extension of $\mathbb{Q}$ is the degree three subfield of $k(y^{1/3})$. The point is that it is obvious from the restriction map that

$$H^1(\mathbb{Q},\mathbb{Z}/3 \mathbb{Z})
= (k^{\times}/k^{\times 3})^{G = {\chi}},$$

where $\chi$ is the non-trivial character of $G = \mathrm{Gal}(\mathbb{Q}(\zeta_3)/\mathbb{Q}) = \mathbb{Z}/2 \mathbb{Z}$.

And basic Kummer theory also says that degree 3 cyclic extensions $K$ of $\mathbb{Q}$ have the form $K(\zeta_3) = \mathbb{Q}(\zeta_3)(\alpha^{1/3})$ where

$$\sigma \alpha = \alpha^{-1} \mod k^{\times}/k^{\times 3}.$$

The same basic structure holds <i>mutatis mutandis</i> with $\mathbb{Q}$ replaced by any number field $F$, and $3$ replaced by $p$, and $G = \chi$ where now $\chi$ is the mod-p cyclotomic character of $G = \mathrm{Gal}(F(\zeta_p)/F)$, which is the canonical (possibly trivial) map $G \rightarrow (\mathbb{Z}/p \mathbb{Z})^{\times}$. And now the map from $k^{\times}/k^{\times p}$ is just the projection to the $\chi$-eigenspace.