1. **Stating the problem:** Given $x > 0$ and $$\frac{x^2}{1 + x^4} = \frac{1}{3},$$ find the value of $$\frac{x^4}{1 + x^8}.$$\n\n2. **Use the given equation:** From $$\frac{x^2}{1 + x^4} = \frac{1}{3},$$ cross-multiply to get $$3x^2 = 1 + x^4.$$\n\n3. **Rewrite the equation:** Rearranged, $$x^4 - 3x^2 + 1 = 0.$$\n\n4. **Substitute:** Let $$y = x^2,$$ so the equation becomes $$y^2 - 3y + 1 = 0.$$\n\n5. **Solve quadratic for y:** Using the quadratic formula, $$y = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}.$$\nSince $x > 0$, $y = x^2 > 0$, so both roots are positive.\n\n6. **Recall the original expression to find:** $$\frac{x^4}{1 + x^8} = \frac{y^2}{1 + y^4}.$$\n\n7. **Use the original relation to find a simpler form:** From step 2, $$3x^2 = 1 + x^4 \Rightarrow 3y = 1 + y^2.$$\nRearranged, $$y^2 = 3y - 1.$$\n\n8. **Calculate numerator:** $$y^2 = 3y - 1.$$\n\n9. **Calculate denominator:** $$1 + y^4 = 1 + (y^2)^2 = 1 + (3y - 1)^2 = 1 + (9y^2 - 6y + 1) = 1 + 9y^2 - 6y + 1 = 9y^2 - 6y + 2.$$\n\n10. **Substitute $y^2$ again:** Replace $y^2$ with $3y - 1$ in denominator:\n$$9y^2 - 6y + 2 = 9(3y - 1) - 6y + 2 = 27y - 9 - 6y + 2 = 21y - 7.$$\n\n11. **Final expression:** $$\frac{y^2}{1 + y^4} = \frac{3y - 1}{21y - 7} = \frac{3y - 1}{7(3y - 1)} = \frac{1}{7}.$$\n\n**Answer:** $$\boxed{\frac{1}{7}}.$$