Let $w^{R}: G \rightarrow \mathfrak{g}\otimes \mathfrak{g}$ be the right translate of the Poisson bivector $w$ of $G$ to the identity, and let $\delta : \mathfrak{g}\rightarrow \mathfrak{g} \otimes \mathfrak{g}$ be the tangent linear map of $w^{R}$ at $e$. Then, we have \begin{align} [\xi_{1},\xi_{2}]_{\mathfrak{g}^{\star}}=\delta^{\star}(\xi_{1}\otimes \xi_{2}), \end{align} where, $\mathfrak{g}$ is a Lie algebra, and $\mathfrak{g}^{\star}$ its dual Lie algebra, and $\xi_{1},\xi_{2} \in \mathfrak{g}^{\star}$. For any $g\in {G}$, \begin{align} \{f_{1},f_{2}\}(g)=\langle w^{R}(g),((R_{g})^{'}_{e}\otimes {(R_{g})^{'}_{e}})^{\star}((df_{1})_{g}\otimes (df_{2})_{g})\rangle. \end{align} Differentiating this equation at $g=e$ in the direction $X\in \mathfrak{g}$, and using the fact that $w^{R}(e)=0$, gives \begin{align} \langle X, d \{f_{1},f_{2}\}_{e}\rangle=\langle \delta (X),\xi_{1} \otimes \xi_{2} \rangle. \end{align}

My question is how to differentiate the second eauation and how to proof the last equation.

Let $w^{R}: G \rightarrow \mathfrak{g}\otimes \mathfrak{g}$ be the right translate of the Poisson bivector $w$ of $G$ to the identity, and let $\delta : \mathfrak{g}\rightarrow \mathfrak{g} \otimes \mathfrak{g}$ be the tangent linear map of $w^{R}$ at $e$. Then, we have \begin{align} [\xi_{1},\xi_{2}]_{\mathfrak{g}^{\star}}=\delta^{\star}(\xi_{1}\otimes \xi_{2}), \end{align} where, $\mathfrak{g}$ is a Lie algebra, and $\mathfrak{g}^{\star}$ its dual Lie algebra, and $\xi_{1},\xi_{2} \in \mathfrak{g}^{\star}$. For any $g\in {G}$, \begin{align} \{f_{1},f_{2}\}(g)=\langle w^{R}(g),((R_{g})^{'}_{e}\otimes {(R_{g})^{'}_{e}})^{\star}((df_{1})_{g}\otimes (df_{2})_{g})\rangle. \end{align} Differentiating this equation at $g=e$ in the direction $X\in \mathfrak{g}$, and using the fact that $w^{R}(e)=0$, gives \begin{align} \langle X, d \{f_{1},f_{2}\}_{e}\rangle=\langle \delta (X),\xi_{1} \otimes \xi_{2} \rangle. \end{align}

My question is how to differentiate the second eauation and how to proof the last equation?