1. Stating the problem: Find the limit $$\lim_{x \to 0} \left(\sqrt{x^2 + 9} - 3\right) / x^2.$$\n\n2. Formula and rules: When direct substitution leads to an indeterminate form like $$0/0$$, we use algebraic manipulation such as rationalizing the numerator.\n\n3. Multiply numerator and denominator by the conjugate of the numerator: $$\frac{\sqrt{x^2 + 9} - 3}{x^2} \times \frac{\sqrt{x^2 + 9} + 3}{\sqrt{x^2 + 9} + 3} = \frac{(\sqrt{x^2 + 9})^2 - 3^2}{x^2 (\sqrt{x^2 + 9} + 3)}.$$\n\n4. Simplify numerator: $$x^2 + 9 - 9 = x^2.$$\n\n5. Substitute back: $$\frac{x^2}{x^2 (\sqrt{x^2 + 9} + 3)}.$$\n\n6. Cancel common factor $$x^2$$: $$\frac{\cancel{x^2}}{\cancel{x^2} (\sqrt{x^2 + 9} + 3)} = \frac{1}{\sqrt{x^2 + 9} + 3}.$$\n\n7. Take the limit as $$x \to 0$$: $$\lim_{x \to 0} \frac{1}{\sqrt{x^2 + 9} + 3} = \frac{1}{\sqrt{0 + 9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}.$$\n\nFinal answer: $$\boxed{\frac{1}{6}}.$$