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.
