1. **State the problem:** Evaluate the triple integral $$\int_0^9 \int_0^{\frac{y}{3}} \int_0^{\sqrt{y^2 - 9x^2}} z \, dz \, dx \, dy.$$\n\n2. **Recall the formula for integrating with respect to $z$: **\n$$\int_0^{\sqrt{y^2 - 9x^2}} z \, dz = \left[ \frac{z^2}{2} \right]_0^{\sqrt{y^2 - 9x^2}} = \frac{y^2 - 9x^2}{2}.$$\n\n3. **Substitute this result into the integral:**\n$$\int_0^9 \int_0^{\frac{y}{3}} \frac{y^2 - 9x^2}{2} \, dx \, dy.$$\n\n4. **Integrate with respect to $x$: **\n$$\int_0^{\frac{y}{3}} \frac{y^2 - 9x^2}{2} \, dx = \frac{1}{2} \int_0^{\frac{y}{3}} (y^2 - 9x^2) \, dx = \frac{1}{2} \left[ y^2 x - 3x^3 \right]_0^{\frac{y}{3}}.$$\n\n5. **Evaluate the expression at the limits:**\n$$\frac{1}{2} \left( y^2 \cdot \frac{y}{3} - 3 \left( \frac{y}{3} \right)^3 \right) = \frac{1}{2} \left( \frac{y^3}{3} - 3 \cdot \frac{y^3}{27} \right) = \frac{1}{2} \left( \frac{y^3}{3} - \frac{y^3}{9} \right).$$\n\n6. **Simplify inside the parentheses:**\n$$\frac{y^3}{3} - \frac{y^3}{9} = \frac{3y^3}{9} - \frac{y^3}{9} = \frac{2y^3}{9}.$$\n\n7. **Multiply by $\frac{1}{2}$:**\n$$\frac{1}{2} \cdot \frac{2y^3}{9} = \frac{y^3}{9}.$$\n\n8. **Now the integral reduces to:**\n$$\int_0^9 \frac{y^3}{9} \, dy = \frac{1}{9} \int_0^9 y^3 \, dy.$$\n\n9. **Integrate with respect to $y$: **\n$$\int_0^9 y^3 \, dy = \left[ \frac{y^4}{4} \right]_0^9 = \frac{9^4}{4} = \frac{6561}{4}.$$\n\n10. **Multiply by $\frac{1}{9}$:**\n$$\frac{1}{9} \cdot \frac{6561}{4} = \frac{6561}{36} = 182.25.$$\n\n**Final answer:** $$\boxed{182.25}.$$