1. **State the problem:** We want to find the value of $$\frac{x^2}{x^4 + 1}$$ given that $$\frac{x^2}{x^2 + 1} = \frac{1}{4}$$. 2. **Given equation:** $$\frac{x^2}{x^2 + 1} = \frac{1}{4}$$. 3. **Solve for $x^2$:** Multiply both sides by $$x^2 + 1$$: $$x^2 = \frac{1}{4}(x^2 + 1)$$ 4. Distribute $$\frac{1}{4}$$: $$x^2 = \frac{1}{4}x^2 + \frac{1}{4}$$ 5. Subtract $$\frac{1}{4}x^2$$ from both sides: $$x^2 - \frac{1}{4}x^2 = \frac{1}{4}$$ 6. Simplify left side: $$\frac{3}{4}x^2 = \frac{1}{4}$$ 7. Multiply both sides by $$\frac{4}{3}$$: $$x^2 = \frac{1}{3}$$ 8. **Find $$x^4$$:** Square $$x^2$$: $$x^4 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$$ 9. **Calculate the desired expression:** $$\frac{x^2}{x^4 + 1} = \frac{\frac{1}{3}}{\frac{1}{9} + 1} = \frac{\frac{1}{3}}{\frac{10}{9}} = \frac{1}{3} \times \frac{9}{10} = \frac{3}{10}$$ **Final answer:** $$\frac{x^2}{x^4 + 1} = \frac{3}{10}$$