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
$$f\left( x \right) =f\left( 0 \right) +{ f\left( 0 \right) }^{ \prime }x+\frac { { f\left( 0 \right) }^{ \prime \prime } }{ 2! } { x }^{ 2 }+\frac { { f\left( 0 \right) }^{ \prime \prime \prime } }{ 3! } { x }^{ 3 }+\frac { { f\left( 0 \right) }^{ \prime \prime \prime \prime } }{ 4! } { x }^{ 4 }+...\\ f\left( x \right) =\cos ^{ 6 }{ x } \\ f\left( 0 \right) =1\\ f^{ \prime }\left( x \right) =-6\cos ^{ 5 }{ x } \sin { x } \Rightarrow f^{ \prime }\left( 0 \right) =-6\cos ^{ 5 }{ 0 } \sin { 0 } =0\\ f^{ \prime \prime }\left( x \right) ={ \left( -6\cos ^{ 5 }{ x } \sin { x } \right) }^{ \prime }=30\cos ^{ 4 }{ x } \sin ^{ 2 }{ x-6\cos ^{ 6 }{ x } } \Rightarrow f^{ \prime \prime }\left( 0 \right) =-6\\ f^{ \prime \prime \prime }\left( x \right) ={ \left( 30\cos ^{ 4 }{ x } \sin ^{ 2 }{ x-6\cos ^{ 6 }{ x } } \right) }^{ \prime }=-120\cos ^{ 3 }{ x } \sin ^{ 3 }{ x } +60\cos ^{ 5 }{ x } \sin { x } +36\cos ^{ 5 }{ x } \sin { x } \Rightarrow f^{ \prime \prime \prime }\left( 0 \right) =0\\ f^{ \prime \prime \prime \prime }\left( x \right) ={ \left( -120\cos ^{ 3 }{ x } \sin ^{ 3 }{ x } +60\cos ^{ 5 }{ x } \sin { x } +36\cos ^{ 5 }{ x } \sin { x } \right) }^{ \prime }=360\cos ^{ 2 }{ x } \sin ^{ 4 }{ x } -360\sin ^{ 2 }{ x } \cos ^{ 4 }{ x } -300\cos ^{ 4 }{ x } \sin ^{ 2 }{ x } +60\cos ^{ 6 }{ x } -180\cos ^{ 4 }{ x } \sin ^{ 2 }{ x } +36\cos ^{ 5 }{ x } \Rightarrow \\ f^{ \prime \prime \prime \prime }\left( 0 \right) =96\\ \cos ^{ 6 }{ x } =1-\frac { 6 }{ 2! } { x }^{ 2 }+\frac { 96 }{ 4! } { x }^{ 4 }+...=1-3{ x }^{ 2 }+4{ x }^{ 4 }+..\\ $$