1. **State the problem:** Evaluate the definite integral $$\int_0^{3\sqrt{2}} \frac{dx}{(36 - x^2)^{3/2}}.$$\n\n2. **Recall the formula:** The integral $$\int \frac{dx}{(a^2 - x^2)^{3/2}} = \frac{x}{a^2 \sqrt{a^2 - x^2}} + C,$$ where $a$ is a constant.\n\n3. **Identify $a$:** Here, $a^2 = 36$, so $a = 6$.\n\n4. **Apply the formula to the definite integral:**\n$$\int_0^{3\sqrt{2}} \frac{dx}{(36 - x^2)^{3/2}} = \left[ \frac{x}{36 \sqrt{36 - x^2}} \right]_0^{3\sqrt{2}}.$$\n\n5. **Evaluate the expression at the upper limit $x = 3\sqrt{2}$:**\nCalculate the denominator inside the square root:\n$$36 - (3\sqrt{2})^2 = 36 - 9 \times 2 = 36 - 18 = 18.$$\nSo,\n$$\frac{3\sqrt{2}}{36 \sqrt{18}}.$$\nSimplify $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$, so\n$$\frac{3\sqrt{2}}{36 \times 3\sqrt{2}} = \frac{3\sqrt{2}}{108 \sqrt{2}}.$$\nCancel $\sqrt{2}$:\n$$\frac{3\cancel{\sqrt{2}}}{108 \times 3 \cancel{\sqrt{2}}} = \frac{3}{108 \times 3} = \frac{3}{324}.$$\nSimplify $\frac{3}{324} = \frac{1}{108}$.\n\n6. **Evaluate the expression at the lower limit $x=0$:**\n$$\frac{0}{36 \sqrt{36 - 0}} = 0.$$\n\n7. **Subtract lower limit from upper limit:**\n$$\frac{1}{108} - 0 = \frac{1}{108}.$$\n\n**Final answer:** $$\boxed{\frac{1}{108}}.$$