1. **State the problem:** Find the vertical asymptotes of the function $$f(x) = \frac{x}{x^2 - 36}$$. 2. **Recall the rule for vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and the numerator is not zero at those points. 3. **Set the denominator equal to zero:** $$x^2 - 36 = 0$$ 4. **Solve for $x$:** $$x^2 = 36$$ $$x = \pm 6$$ 5. **Check the numerator at these points:** At $x=6$, numerator is $6 \neq 0$. At $x=-6$, numerator is $-6 \neq 0$. 6. **Conclusion:** Vertical asymptotes are at $x=6$ and $x=-6$. **Final answer:** $$x=6 \text{ and } x=-6$$