1. **State the problem:** We need to evaluate the integral $$\int \frac{8x^2}{(x^3+3)^3} \, dx.$$\n\n2. **Identify a substitution:** Notice the denominator has $(x^3+3)^3$ and the numerator has $8x^2$. The derivative of $x^3+3$ is $3x^2$, which is similar to the numerator. This suggests the substitution $ u = x^3 + 3 $.\n\n3. **Compute differential:** \n$$du = 3x^2 \, dx \implies x^2 \, dx = \frac{du}{3}.$$\n\n4. **Rewrite the integral in terms of $u$:**\n$$\int \frac{8x^2}{(x^3+3)^3} \, dx = \int \frac{8}{u^3} \cdot x^2 \, dx = \int \frac{8}{u^3} \cdot \frac{du}{3} = \int \frac{8}{3u^3} \, du.$$\n\n5. **Simplify the integral:**\n$$\int \frac{8}{3u^3} \, du = \frac{8}{3} \int u^{-3} \, du.$$\n\n6. **Integrate:**\n$$\int u^{-3} \, du = \frac{u^{-2}}{-2} + C = -\frac{1}{2u^2} + C.$$\n\n7. **Substitute back to $x$:**\n$$\frac{8}{3} \cdot \left(-\frac{1}{2u^2}\right) + C = -\frac{8}{6u^2} + C = -\frac{4}{3(x^3+3)^2} + C.$$\n\n**Final answer:**\n$$\boxed{-\frac{4}{3(x^3+3)^2} + C}.$$