Start with the known Maclaurin series of \(\displaystyle \cos x \)

\(\displaystyle \cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + ... \)

then \(\displaystyle \cos 6x^{2} = 1 - \frac{(6x^{2})^{2}}{2!} + \frac{(6x^{2})^{4}}{4!} - \frac{(6x^{2})^{6}}{6!} + ... = 1 - \frac{6^{2}x^{4}}{2!} + \frac{6^{4}x^{8}}{4!} - \frac{6^{6}x^{12}}{6!} + ... \)

and \(\displaystyle \cos 6x^{2} -1 = -\frac{6^{2}x^{4}}{2!} + \frac{6^{4}x^{8}}{4!} - \frac{6^{6}x^{12}}{6!} + ...\)

finally \(\displaystyle \frac{\cos 6x^{2}-1}{x^{2}} = -\frac{6^{2}x^{2}}{2!} + \frac{6^{4}x^{6}}{4!} - \frac{6^{6}x^{10}}{6!} + ...\)

For a Macluarin series, the coefficent of the \(\displaystyle x^{10} \) term is \(\displaystyle \frac{f^{(10)}(0)}{10!} \)

The coefficient of the \(\displaystyle x^{10} \) term in the above Macluarin series is \(\displaystyle -\frac{6^{6}}{6!} \)

so \(\displaystyle \frac{f^{(10)}(0)}{10!} = -\frac{6^{6}}{6!} \)

EDIT: and \(\displaystyle f^{(10)}(0) = - \frac{6^{6} 10!}{6!} = - 235146240\) (Surprised)