1. **State the problem:** Given the system of equations: $$x^2 = 2025x + y$$ $$y^2 = 2025y + x$$ Find the value of $$\frac{x}{y^2} + \frac{y}{x^2}$$. 2. **Rewrite the equations:** From the first equation: $$x^2 - 2025x = y$$ From the second equation: $$y^2 - 2025y = x$$ 3. **Substitute and analyze:** We want to find: $$\frac{x}{y^2} + \frac{y}{x^2}$$ Rewrite as: $$\frac{x}{y^2} + \frac{y}{x^2} = \frac{x^3 + y^3}{x^2 y^2}$$ 4. **Express $x^3$ and $y^3$ using the original equations:** Multiply the first equation by $x$: $$x^3 = 2025x^2 + xy$$ Multiply the second equation by $y$: $$y^3 = 2025y^2 + xy$$ 5. **Sum $x^3 + y^3$:** $$x^3 + y^3 = 2025(x^2 + y^2) + 2xy$$ 6. **Express denominator $x^2 y^2$ as $(xy)^2$:** 7. **Use the original equations to find relations between $x$ and $y$:** From the first equation: $$y = x^2 - 2025x$$ From the second equation: $$x = y^2 - 2025y$$ 8. **Try the symmetric solution $x = y$:** Set $x = y = t$: $$t^2 = 2025t + t \implies t^2 - 2026t = 0 \implies t(t - 2026) = 0$$ So $t = 0$ or $t = 2026$. 9. **Check $t=0$:** Plug into original equations: $$0 = 0 + 0$$ Valid. 10. **Check $t=2026$:** $$2026^2 = 2025 \times 2026 + 2026$$ Calculate: $$2025 \times 2026 = 2025 \times (2000 + 26) = 2025 \times 2000 + 2025 \times 26 = 4,050,000 + 52,650 = 4,102,650$$ Then: $$2025 \times 2026 + 2026 = 4,102,650 + 2026 = 4,104,676$$ Calculate $2026^2$: $$2026^2 = (2000 + 26)^2 = 2000^2 + 2 \times 2000 \times 26 + 26^2 = 4,000,000 + 104,000 + 676 = 4,104,676$$ So equality holds. 11. **Evaluate the expression for $x = y = 0$:** $$\frac{0}{0^2} + \frac{0}{0^2}$$ Undefined (division by zero). 12. **Evaluate the expression for $x = y = 2026$:** $$\frac{2026}{2026^2} + \frac{2026}{2026^2} = \frac{2026}{2026^2} + \frac{2026}{2026^2} = 2 \times \frac{2026}{2026^2} = 2 \times \frac{1}{2026} = \frac{2}{2026}$$ Simplify: $$\frac{2}{2026} = \frac{1}{1013}$$ 13. **Final answer:** $$\boxed{\frac{1}{1013}}$$