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
Let's write
$$1 = \cos{x} \sec{x} = \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots\right)(a_0 + a_1 x + a_2x^2 + \dots)$$
Expand the right hand side to find that
$$1 = 1 (a_0) + x (a_1) + x^2 \left(a_2 - a_0 \frac{1}{2!}\right) + x^3 \left(a_3 - a_1 \frac{1}{2!}\right) + x^4 \left(a_4 - a_2 \frac{1}{2!} + a_0 \frac{1}{4!}\right)+\dots$$
Equating coefficients gives (note that the left side is $1 + 0x + 0x^2 + 0x^3 + \dots$)
\begin{align*} 1 &= a_0 \\ 0 &= a_1 \\ 0 &= a_2 - \frac{a_0}{2} \\ 0 &= a_3 - \frac{a_1}{2} \\ 0 &= a_4 - \frac{a_2}{2} + \frac{a_0}{24} \end{align*}
Solving this gives $a_0 = 1$, $a_1 = 0 = a_3$, $a_2 = \frac{1}{2}$ and $a_4 = \frac{5}{24}$, or
$$\boxed{\sec{x} \approx 1 + \frac{1}{2} x^2 + \frac{5}{24} x^4}$$