Infinitesimal Generator of a Group Action This question is from IV.3, Exercise 6 in Boothby's Introduction to Differentiable Manifolds.  Given the action $$\theta(t; x, y) = (xe^{2t}, ye^{-3t})$$ from $\mathbb{R}\times\mathbb{R}^2 \rightarrow \mathbb{R}^2$ (treating $\mathbb{R}$ as your group and $\mathbb{R}^2$ as the manifold), determine the infinitesimal generator $X_p$ (herein "inf gen") and show that it is $\theta$-invariant.
He generically defines the infinitesimal generator as $$X_p f = \sum^n_{i=1} \dot{h^i}(0,x)\left(\frac{\partial \hat{f}}{\partial x^i}\right)_{\phi(p)}.$$
Here $x^i = {h^i}(0,x)$ and $\hat{f}(x^1, \ldots, x^n)$ is a local expression for the function $f\in C^\infty$.  The overdot indicates a derivative with respect to $t$.
Thank you in advance for any help.
[I originally asked for hints and geometric intuition.  See below for the solution attempt.]
 A: The answer per the hint is given as follows.
We have $\theta(t; x, y) = (xe^{2t}, ye^{-3t})$.  On inspection, we see that it is an action, as $\theta(0; x, y) = (x, y)$ and $\theta(t+s; x, y) = \theta(s+t; x, y)$.  Per the note, we compute
\begin{equation} \label{eq1}
\begin{split}
\dot{\theta_1} & = 2xe^{2t} \\
        & = 2\theta_1 \\
\dot{\theta_2} & = -3ye^{-3t} \\
        & = -3\theta_2. \\
\end{split}
\end{equation}
This means that the infinitesimal generator $X_{(x,y)} = 2x \frac{\partial}{\partial x} -3y \frac{\partial}{\partial y}$.  By Theorem 3.4 in Boothby IV.3, $X$ is $\theta$-invariant.  Without citing this, we can check if $X$ is $\theta$-invariant, i.e. if $\theta_{t*}(x,y) = X_{\theta_(x,y)}$.
The left-hand side has $X_{(x,y)}$ as above, with $\theta_t(x,y) = (xe^{2t}, ye^{-3t})$, which implies that $X_{\theta_t(x,y)} = 2xe^{2t}\frac{\partial}{\partial x} - 3ye^{-3t}\frac{\partial}{\partial y}$.
The right-hand side requires the computation of the pushforward $\theta_{t*}$.  We consider this $\theta_{t}$ as a function of $x$ and $y$.  Generally, for any $v \in T_pM$ with $v = v_1 \frac{\partial}{\partial x} +v_2 \frac{\partial}{\partial y}$, $$\theta_{t*} (v) =
\begin{pmatrix}
    \frac{\partial \theta_{t,1}}{\partial x}  &  \frac{\partial \theta_{t,1}}{\partial y}      \\
    \frac{\partial \theta_{t,2}}{\partial x}  &  \frac{\partial \theta_{t,2}}{\partial y}   
\end{pmatrix}
\begin{pmatrix}
    v_1 \\
    v_2
\end{pmatrix} 
$$
where the basis for $v$ (the matrix on the right) is implicitly $(\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$.  Computing the Jacobian, we have
$$\theta_{t*} =
\begin{pmatrix}
   e^{2t}  &  0    \\
   0  &  e^{-3t}  
\end{pmatrix}.$$
So $\theta_{t*}(X_{(x,y)}) = \theta_{t*}(2x \frac{\partial}{\partial x} -3y \frac{\partial}{\partial y})$, and substituting into the above, we have that
$$\theta_{t*}(X_{(x,y)}) =
\begin{pmatrix}
   e^{2t}  &  0    \\
   0  &  e^{-3t}  
\end{pmatrix}
\begin{pmatrix}
    2x \\
    -3y
\end{pmatrix}
=
\begin{pmatrix}
    2xe^{2t} \\
    -3ye^{-3y}
\end{pmatrix} .$$
Here again the basis is understood to be $(\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$.  So we are done.
