I'm reading the well-known article of Reyman and Semenov-Tian-Shansky called "Group-Theoretical Methods in the Theory of Finite-Dimensional Integrable Systems".
Suppose $\mathfrak{g}$ is a Lie algebra, and $R$ is an $R$-matrix (so $[x,y]_R = [Rx,y] + [x,Ry]$ is a Lie bracket on $\mathfrak{g}$). So $\mathfrak{g}$ has two Lie brackets (the usual one, and $[-,-]_R$), and thus $\mathfrak{g}^*$ has two Lie-Poisson structures. I'm reading the proof of the fact that if $f$ and $g$ are coadjoint-invariant functions on $\mathfrak{g}^*$, then $\{f,g\}_R = 0$.
The authors say that
"The invariance of $f$ is equivalent to the relation $\left<\lambda, [d_\lambda f, x] \right> = 0$ for all $x \in \mathfrak{g}$"
Here, $df$ (at a point $\lambda \in \mathfrak{g}^*$) is identified with an element of $\mathfrak{g}$, and $\left<-,-\right>$ is the dual pairing of $\mathfrak{g}^*$ and $\mathfrak{g}$.
This is their entire proof of this fact. I see why the result follows from this, but I do not see why this statement is true. Why is the fact that $f$ is coadjoint-invariant equivalent to this relation?
Let $G$ be a simply connected Lie group with Lie algebra $\mathfrak{g}$. Note that $f$ being coadjoint invariant means that \begin{align*} f(\mathrm{Ad}^*_g(\lambda))= f(\lambda), \quad \forall \lambda \in \mathfrak{g}^*,g\in G \end{align*} which for $g_t=e^{tx}$ with $x\in \mathfrak{g}$ implies \begin{align*} \frac{d}{dt}|_{t=0}f(\mathrm{Ad}^*_{e^tx}(\lambda))= \frac{d}{dt}|_{t=0} f(\lambda) =0 \end{align*} but also \begin{align*} \frac{d}{dt}|_{t=0}f(\mathrm{Ad}^*_{e^tx}(\lambda))= \mathrm{d}_{\lambda}f(\mathrm{ad}^*_x(\lambda)) =\lambda ( [\mathrm{d}_{\lambda}f,x]). \end{align*}