I'm studying Poisson geometry from "Lectures on Poisson Geometry" (Crainic, Fernandes, Marcut). Let $M$ be a Poisson manifold and let $X$ be a vector field on $M$. We define $X$ to be Poisson iff $X$ is a derivation of the Lie algebra $(C^\infty(M),\{-,-\})$: $$X(\{f,g\})=\{X(f),g\}+\{f,X(g)\}$$ I'm having trouble with the following exercise
Prove that $X$ is Poisson iff its flow $\Phi^t_X$ is a Poisson diffeomorphism.
My attempt By definition, the fact that $\Phi^t_X$ is Poisson means that: $$\{f\circ \Phi^t_X,g\circ \Phi^t_X\}-\{f,g\}\circ \Phi^t_X=0$$ This implies that the derivative with respect to $t$ of the expression above is $0$ $$\frac{d}{dt}|_{t=\overline{t}} \{f\circ \Phi^t_X,g\circ \Phi^t_X\}-\{f,g\}\circ \Phi^t_X=$$ $$=\frac{d}{dt}|_{t=0} \{f\circ \Phi^t_X \circ \Phi^{\overline{t}}_X,g\circ \Phi^t_X \circ \Phi^{\overline{t}}_X\}-\{f,g\}\circ \Phi^t_X\circ \Phi^{\overline{t}}=$$ $$=\left(\frac{d}{dt}|_{t=0} \{f\circ \Phi^t_X \circ \Phi^{\overline{t}}_X,g\circ \Phi^t_X \circ \Phi^{\overline{t}}_X\}\right)-X(\{f,g\})\circ \Phi^{\overline{t}}_X$$
Applying Leibnitz rule to the bilinear form $\{-,-\}$, I get:
$$=\{X(f)\circ \Phi^{\overline{t}}_X, g\circ \Phi^{\overline{t}}_X\}+\{f\circ \Phi^{\overline{t}}_X,X(g)\circ \Phi^{\overline{t}}_X\}+X(\{f,g\})\circ \Phi^{\overline{t}}_X$$
By imposing $\overline{t}=0$ we obtain that $X$ is Poisson, so one implication is done. But now I don't manage to prove that if $X$ is Poisson then its flow is a Poisson diffeomorphism.
(I'm aware that this is obvious for Hamiltonian fields since the flow being a Poisson diffeomorphism simply translates to the Jacobi identity, but this broader case seems harder).
This may not be exactly what you're looking for, but this problem is more easily solved using Poisson bivector fields. Let $\pi\in\Gamma(\Lambda^2TM)$ be the Poisson bivector field associated to $\{\ ,\ \}$, i.e. the unique bivector field satisfying $\{f,g\}=\langle\pi,df\wedge dg\rangle$.
To rephrase things in terms of a $\pi$, we say a vector field $X\in\mathfrak{X}M$ is Poisson if $\mathcal{L}_X\pi=0$, and that a diffeomorphism $\varphi:M\to M$ is Poisson if $\varphi^*\pi=\pi$. It may be an instructive exercise to show that these definitions are equivalent to the ones using brackets.
From here, the proof is the same as is used for Killing vector fields and other "structure-preserving flows". Let $\pi_t=(\Phi_X^t)^*\pi$. We can rewrite the ODE for $\pi_t$ using some dummy variables. $$\begin{align} \frac{d}{dt}\pi_t&=\frac{d}{d\tau}\left[\left(\Phi_X^\tau\right)^*\pi\right]_{\tau=t} \\ &=\frac{d}{d\tau}\left[\left(\Phi_X^{t+\tau}\right)^*\pi\right]_{\tau=0} \\ &=\frac{d}{d\tau}\left[\left(\Phi_X^t\right)^*\left(\Phi_X^\tau\right)^*\pi\right]_{\tau=0} \\ &=\left(\Phi_X^t\right)^*\frac{d}{d\tau}\left[\left(\Phi_X^\tau\right)^*\pi\right]_{\tau=0} \\ &=\left(\Phi_X^t\right)^*\left(\mathcal{L}_X\pi\right) \\ \end{align}$$ If $X$ is Poisson, the right side vanishes for all $t$, so the ODE clearly has only constant solutions and thus $\Phi_X^t$ is Poisson.
Edit:
All the bivector formulation is doing is organizing the above computation in a covenient way. To do things in terms of brackets, we can simply translate it line by line, at some cost in transparency. Let $\{f,g\}_{t}=\{f\circ\Phi_X^{-t},g\circ\Phi_X^{-t}\}\circ\Phi_X^t$. $$ \frac{d}{dt}\{f,g\}_{t}=\frac{d}{d\tau}\left[\{f\circ\Phi_X^{-t}\circ\Phi_X^{-\tau},g\circ\Phi_X^{-t}\circ\Phi_X^{-\tau}\}\circ\Phi_X^\tau\circ\Phi_X^t\right]_{\tau=0} $$ Since the bracket is a bilinear function on $C^\infty M$, we can apply the product and chain rules, yielding $$\begin{align} \frac{d}{dt}\{f,g\}_{t}=&X\left(\{f\circ\Phi_X^{-t},g\circ\Phi_X^{-t}\}\right)\circ\Phi_X^t \\ &-\{X\left(f\circ\Phi_X^{-t}\right),g\circ\Phi_X^{-t}\}\circ\Phi_X^t \\ &-\{f\circ\Phi_X^{-t},X\left(g\circ\Phi_X^{-t}\right)\}\circ\Phi_X^t \end{align}$$ Writing $f_t=f\circ\Phi_X^{-t}$ and $g_t=g\circ\Phi_X^{-t}$, we have $$ \frac{d}{dt}\{f,g\}_{t}=\left[X(\{f_t,g_t\})-\{X(f_t),g_t\}-\{f_t,X(g_t)\}\right]\circ\Phi_X^t $$ And once again, if $X$ is Poisson, the ODE has only constant solutions $\{f,g\}_t=\{f,g\}$ and thus $\Phi_X^t$ is Poisson.