1. **State the problem:** Solve the inequality $$\left(-x - \frac{1}{3}\right)^2 + (3x + 1)^2 - \frac{x}{2} + \frac{1}{6} \geq 10x(x - 1) - \frac{37}{9}.$$\n\n2. **Expand the squares:**\n$$\left(-x - \frac{1}{3}\right)^2 = \left(-x - \frac{1}{3}\right)\left(-x - \frac{1}{3}\right) = x^2 + 2 \cdot x \cdot \frac{1}{3} + \frac{1}{9} = x^2 + \frac{2x}{3} + \frac{1}{9}.$$\n$$ (3x + 1)^2 = 9x^2 + 6x + 1.$$\n\n3. **Substitute expansions back:**\n$$x^2 + \frac{2x}{3} + \frac{1}{9} + 9x^2 + 6x + 1 - \frac{x}{2} + \frac{1}{6} \geq 10x^2 - 10x - \frac{37}{9}.$$\n\n4. **Combine like terms on the left:**\n$$x^2 + 9x^2 = 10x^2,$$\n$$\frac{2x}{3} + 6x - \frac{x}{2} = \frac{2x}{3} + \frac{12x}{2} - \frac{x}{2} = \frac{2x}{3} + \frac{11x}{2},$$\n$$\frac{1}{9} + 1 + \frac{1}{6} = \frac{1}{9} + \frac{9}{9} + \frac{3}{18} = \frac{1}{9} + 1 + \frac{1}{6} = \frac{1}{9} + \frac{9}{9} + \frac{3}{18} = \frac{1}{9} + 1 + \frac{1}{6}.$$\nLet's convert all to a common denominator 18:\n$$\frac{1}{9} = \frac{2}{18}, \quad 1 = \frac{18}{18}, \quad \frac{1}{6} = \frac{3}{18}.$$\nSum: $$\frac{2}{18} + \frac{18}{18} + \frac{3}{18} = \frac{23}{18}.$$\n\n5. **Rewrite the inequality:**\n$$10x^2 + \left(\frac{2x}{3} + 6x - \frac{x}{2}\right) + \frac{23}{18} \geq 10x^2 - 10x - \frac{37}{9}.$$\n\n6. **Simplify the $x$ terms:**\nFind common denominator for $\frac{2x}{3} + 6x - \frac{x}{2}$:\n$$\frac{2x}{3} = \frac{4x}{6}, \quad 6x = \frac{36x}{6}, \quad -\frac{x}{2} = -\frac{3x}{6}.$$\nSum: $$\frac{4x}{6} + \frac{36x}{6} - \frac{3x}{6} = \frac{37x}{6}.$$\n\n7. **Subtract $10x^2$ from both sides:**\n$$10x^2 + \frac{37x}{6} + \frac{23}{18} - 10x^2 \geq 10x^2 - 10x - \frac{37}{9} - 10x^2,$$\nwhich simplifies to\n$$\frac{37x}{6} + \frac{23}{18} \geq -10x - \frac{37}{9}.$$\n\n8. **Bring all terms to left side:**\n$$\frac{37x}{6} + 10x + \frac{23}{18} + \frac{37}{9} \geq 0.$$\n\n9. **Combine $x$ terms:**\n$$\frac{37x}{6} + 10x = \frac{37x}{6} + \frac{60x}{6} = \frac{97x}{6}.$$\n\n10. **Combine constants:**\nConvert $\frac{37}{9}$ to eighteenth denominator: $$\frac{37}{9} = \frac{74}{18}.$$\nSum constants: $$\frac{23}{18} + \frac{74}{18} = \frac{97}{18}.$$\n\n11. **Inequality is now:**\n$$\frac{97x}{6} + \frac{97}{18} \geq 0.$$\n\n12. **Multiply entire inequality by 18 (positive, so inequality direction stays):**\n$$18 \times \left(\frac{97x}{6} + \frac{97}{18}\right) \geq 0,$$\nwhich is\n$$3 \times 97x + 97 \geq 0,$$\nor\n$$291x + 97 \geq 0.$$\n\n13. **Solve for $x$:**\n$$291x \geq -97,$$\n$$x \geq \frac{-97}{291}.$$\n\n**Final answer:**\n$$\boxed{x \geq -\frac{97}{291}}.$$