1. **State the problem:** Find the limit $$\lim_{x \to 0} \left(\sqrt{x^2 + 9} - 3\right) / x^2.$$\n\n2. **Recall the formula and approach:** When direct substitution leads to an indeterminate form like $0/0$, we can use algebraic manipulation such as multiplying by the conjugate to simplify the expression.\n\n3. **Apply the conjugate:** Multiply numerator and denominator by the conjugate of the numerator $$\sqrt{x^2 + 9} + 3$$ to get\n$$\frac{\sqrt{x^2 + 9} - 3}{x^2} \cdot \frac{\sqrt{x^2 + 9} + 3}{\sqrt{x^2 + 9} + 3} = \frac{(\sqrt{x^2 + 9})^2 - 3^2}{x^2 (\sqrt{x^2 + 9} + 3)}.$$\n\n4. **Simplify numerator:**\n$$\frac{x^2 + 9 - 9}{x^2 (\sqrt{x^2 + 9} + 3)} = \frac{x^2}{x^2 (\sqrt{x^2 + 9} + 3)}.$$\n\n5. **Cancel common factors:**\n$$\frac{\cancel{x^2}}{\cancel{x^2} (\sqrt{x^2 + 9} + 3)} = \frac{1}{\sqrt{x^2 + 9} + 3}.$$\n\n6. **Evaluate the limit as $x \to 0$:**\n$$\lim_{x \to 0} \frac{1}{\sqrt{x^2 + 9} + 3} = \frac{1}{\sqrt{0 + 9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}.$$\n\n**Final answer:** $$\boxed{\frac{1}{6}}.$$