Given the function $f(x)=\left(-8 x^{3}-4-8 x^{-2}\right)\left(-9 x^{3}+5\right)$, find $f^{\prime}(x)$ in any form.

Step 1 ：Let's denote \(u(x) = -8 x^{3}-4-8 x^{-2}\) and \(v(x) = -9 x^{3}+5\).

Step 2 ：First, we need to find the derivatives of \(u(x)\) and \(v(x)\).

Step 3 ：The derivative of \(u(x)\) is: \(u'(x) = -24x^{2}+16x^{-3}\).

Step 4 ：The derivative of \(v(x)\) is: \(v'(x) = -27x^{2}\).

Step 5 ：Now, we can apply the product rule: \(f'(x) = u'(x)v(x) + u(x)v'(x)\).

Step 6 ：Substituting the expressions we found earlier, we get: \(f'(x) = (-24x^{2}+16x^{-3})(-9 x^{3}+5) + (-8 x^{3}-4-8 x^{-2})(-27x^{2})\).

Step 7 ：Now, we simplify the expression: \(f'(x) = -216x^{5}+120x^{2}+144x^{-1}-45 + 216x^{5}+108x^{2}+216x^{-2}\).

Step 8 ：Combining like terms, we get: \(f'(x) = 108x^{2}+144x^{-1}-45+216x^{-2}\).

Step 9 ：So, the derivative of the function \(f(x)=\left(-8 x^{3}-4-8 x^{-2}\right)\left(-9 x^{3}+5\right)\) is \(\boxed{f'(x) = 108x^{2}+144x^{-1}-45+216x^{-2}}\).