The Maclaurin expansion for $\sin(x)$ is
$$ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots = \sum_{n = 0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}. \tag{1}$$
In order to get the Maclaurin expansion for $\sin(x^3)$, we plug in $x^3$ into each $x$ appearing in $(1)$,
$$ \sin(x^3) = x^3 - \frac{x^9}{3!} + \frac{x^{15}}{5!} - \cdots = \sum_{n=0}^\infty (-1)^n \frac{x^{6n+3}}{(2n+1)!}.$$
And finally, in order to go from $\sin(x^3)$ to $x^2 \sin(x^3)$, we just multiply each term by $x^2$,
$$ x^2 \sin(x^3) = x^5 - \frac{x^{11}}{3!} + \frac{x^{17}}{5!} - \cdots = \sum_{n = 0}^\infty (-1)^n \frac{x^{6n+5}}{(2n+1)!}.$$
This is the desired expansion. $\diamondsuit$
Here is a tractable way forward to find the coefficients of the series for $\sec(x)$. It is straightforward to adopt this apply this approach to find the series for $9\sec(3x)$.
Recall that $\sec(x)\,\cos(x)=1$ and that the series for the cosine function is given by
$$\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}$$
Furthermore note that since the secant function is even, its Taylor series will be given by
$$\sum_{n=0}^\infty \frac{a_nx^{2n}}{(2n)!}$$
Then, multiplying the series in $(1)$ with the series in $(2)$ yields
$$\begin{align}
1&=\sec(x)\,\cos(x)\\\\
&=\left(\sum_{m=0}^\infty \frac{a_mx^{2m}}{(2m)!}\right)\left(\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}\right)\\\\
&=\sum_{p=0}^\infty\left(\sum_{m=0}^p \frac{(-1)^{p-m}a_{m}}{(2m)!(2(p-m))!}\right)x^{2p}
\end{align}$$
Therefore, we must have
$$\sum_{m=0}^p \frac{(-1)^{p-m}a_{m}}{(2m)!(2(p-m))!}=\begin{cases}
1&,p=0\\\\
0&,p>0
\end{cases} \tag 1$$
We can use $(1)$ to find the coefficients $a_m$ recursively. We see that for $p=0$, $(1)$ reveals that $a_0=1$ and for $p>0$ we have the recursive relationship
$$\bbox[5px,border:2px solid #C0A000]{a_p=-(2p)!\sum_{m=0}^{p-1} \frac{(-1)^{p-m}a_{m}}{(2m)!(2(p-m))!}} \tag 2$$
Let's use $(2)$ to find the first few coefficients of the secant function. For $p=1$, we have
$$a_1=-(2)!\frac{(-1)^{1-0}a_0}{(0)!(2)!}=1$$
For $p=2$, we have
$$a_2=-(4)!\left(\frac{(-1)^{2-0}a_0}{(0)!((4)!)}+\frac{(-1)^{2-1}a_1}{(2)!(2)!}\right)=5$$
For $p=3$, we have
$$a_3=-(6)!\left(\frac{(-1)^{3-0}a_0}{(0)!((6)!)}+\frac{(-1)^{3-1}a_1}{(2)!(4)!}+\frac{(-1)^{3-2}a_2}{(4)!(2)!}\right)=61$$
We can continue recursively to obtain coefficients for higher order terms, but are content here to write the series using the firsts few terms as
$$\bbox[5px,border:2px solid #C0A000]{\sec(x)=1+\frac x2+\frac{5x^2}{24}+\frac{61x^4}{720}+R_8(x)}$$
where the remainder $R_8(x)$ is
$$R_8(x)=\sum_{p=4}^\infty \left(\sum_{m=0}^p \frac{(-1)^{p-m}a_{m}}{(2m)!(2(p-m))!}\right)x^{2p}=O(x^8)
$$
Best Answer
For a positive $R$, this series expansion $$\sin(2x)= \sum_{n=0}^\infty (-1)^n {2^{2n+1}x^{2n+1} \over (2n+1)!}$$ is valid for $x$ in $[-R, R]$. Differentiating term by term gives $$2\cos(2x) = \sum_{n=0}^\infty (-1)^n {2^{2n+1}x^{2n} \over (2n)!},$$ valid for $x$ in $(-R, R)$. Hence this series expansion for $\cos(2x)$ is valid over the whole real line because $R$ was arbitrary.