I'm not terribly sure if this will work in this case, but to extend the idea of expanding the cosine, we can exploit the definition of the Taylor series coefficients. This may be what Xevarion/Dick meant, but by finding the Taylor series of the entire function about a neighbourhood of zero then the coefficient of x^9 will be your solution. I know that this works with much simpler functions, but this one is a bit complicated so I might make an error.
You can get the series representation by expanding the cosine, and treating everything else like polynomials. That is, when you do the expansion, you should get something along the lines of
cos(6x^4) = 1-18x^8+54x^{16} -\ldots
Thus \frac{cos(6x^4) -1 }{x^7} = -18x + 54x^9-\ldots
Now since f(x) = \displaystyle \sum_{n=o}^\infty \frac{f^n(0)}{n!} x^n
then by equation coefficients, you can show that
f^9 (0) = 54\times 9!
Edit: This is the exact same answer you'll get if you just expand the cosine series, since the procedure is precisely the same; however, I find that this is a useful technique that can be used with much nastier functions, so just thought I'd throw in my two-cents
