1. **State the problem:** Find the vertical asymptotes of the function $$f(x) = \frac{x^2 + 5}{(x^2 - 25)(x^2 - 64)}.$$\n\n2. **Recall the rule for vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and the numerator is not zero at those points.\n\n3. **Factor the denominator:**\n$$x^2 - 25 = (x - 5)(x + 5)$$\n$$x^2 - 64 = (x - 8)(x + 8)$$\nSo the denominator is $$(x - 5)(x + 5)(x - 8)(x + 8).$$\n\n4. **Set denominator equal to zero to find potential vertical asymptotes:**\n$$ (x - 5)(x + 5)(x - 8)(x + 8) = 0 $$\nThis gives solutions: $$x = 5, -5, 8, -8.$$\n\n5. **Check numerator at these points:**\n$$x^2 + 5$$ is always positive (since $x^2 \geq 0$ and $+5$), so numerator is never zero at these points.\n\n6. **Conclusion:** All four points are vertical asymptotes.\n\n**Final answer:** The function has four vertical asymptotes. The asymptotes in order from leftmost to rightmost are $$x = -8, x = -5, x = 5, \text{ and } x = 8.$$