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 $G = \mathrm{Gal}(\mathbb{Q}(\zeta_3)/\mathbb{Q}) = \mathbb{Z}/2 \mathbb{Z}$.
(Added Here $M^{\chi}$ means what is says on the tin. If $\sigma \in G$ and $m \in M^{\chi}$, then $\sigma m = \chi(\sigma) m$.)
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 mutatis mutandis 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.
Added: If you want an explicit polynomial, you can, of course, use Galois theory to do so. In fact, everything in this question one can (and I do) teach in the introductory undergraduate Galois theory course. To spell out the elementary details, you want an element of $k(y^{1/3})$ which is fixed by the order two element $\sigma \in \mathrm{Gal}(k(y^{1/3})/\mathbb{Q})$ (there is an obvious splitting from $\mathrm{Gal}(k/\mathbb{Q}) \rightarrow \mathrm{Gal}(k(y^{1/3}/\mathbb{Q})$). The obvious element to take is thus
$$z = y^{1/3} + \sigma y^{1/3} = y^{1/3} + y^{-1/3},$$
which is a root of
$$T^3 - 3 T - (y + y^{-1}) = T^3 - 3 T - \left(\frac{x}{x^{\sigma}} + \frac{x^{\sigma}}{x}\right) = T^3 - 3 T - \frac{Tr(x^2)}{N(x)} \in \mathbb{Q}[T].$$