1. **State the problem:** Find all x-intercepts of the function $$f(x) = \frac{9x^2 - 64}{6x^2 + 16x}$$.\n\n2. **Recall the rule for x-intercepts:** The x-intercepts occur where the function equals zero, i.e., where the numerator is zero and the denominator is not zero.\n\n3. **Set the numerator equal to zero:**\n$$9x^2 - 64 = 0$$\n\n4. **Solve for x:**\n$$9x^2 = 64$$\n$$x^2 = \frac{64}{9}$$\n$$x = \pm \frac{8}{3}$$\n\n5. **Check the denominator at these x-values to ensure it is not zero:**\nDenominator: $$6x^2 + 16x$$\nAt $$x = \frac{8}{3}$$:\n$$6\left(\frac{8}{3}\right)^2 + 16\left(\frac{8}{3}\right) = 6\cdot \frac{64}{9} + \frac{128}{3} = \frac{384}{9} + \frac{384}{9} = \frac{768}{9} \neq 0$$\nAt $$x = -\frac{8}{3}$$:\n$$6\left(-\frac{8}{3}\right)^2 + 16\left(-\frac{8}{3}\right) = 6\cdot \frac{64}{9} - \frac{128}{3} = \frac{384}{9} - \frac{384}{9} = 0$$\n\nSince the denominator is zero at $$x = -\frac{8}{3}$$, this is a vertical asymptote, not an x-intercept.\n\n6. **Therefore, the only x-intercept is at:**\n$$\left(\frac{8}{3}, 0\right)$$\n\n**Final answer:** The function has one x-intercept at $$\boxed{\left(\frac{8}{3}, 0\right)}$$.