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 bounds:**\n- At $x = 3\sqrt{2}$:\n$$\sqrt{36 - (3\sqrt{2})^2} = \sqrt{36 - 18} = \sqrt{18} = 3\sqrt{2}.$$\nSo, numerator: $3\sqrt{2}$\nDenominator: $36 \times 3\sqrt{2} = 108\sqrt{2}$.\nTherefore,\n$$\frac{3\sqrt{2}}{108\sqrt{2}} = \frac{3}{108} = \frac{1}{36}.$$\n- At $x=0$, the expression is $0$.\n\n6. **Subtract to find the definite integral value:**\n$$\frac{1}{36} - 0 = \frac{1}{36}.$$\n\n**Final answer:** $$\int_0^{3\sqrt{2}} \frac{dx}{(36 - x^2)^{3/2}} = \frac{1}{36}.$$