1. **State the problem:** Find the limit as $x$ approaches $-\infty$ of the expression $$\frac{3x+5}{\sqrt{9x^2+2x+6}}.$$\n\n2. **Recall the formula and rules:** When dealing with limits involving radicals and polynomials as $x \to \pm \infty$, we focus on the highest degree terms because they dominate the behavior.\n\n3. **Rewrite the expression focusing on dominant terms:**\n$$\lim_{x \to -\infty} \frac{3x+5}{\sqrt{9x^2+2x+6}} = \lim_{x \to -\infty} \frac{3x+5}{\sqrt{9x^2(1 + \frac{2}{9x} + \frac{6}{9x^2})}}.$$\n\n4. **Simplify the square root:**\n$$= \lim_{x \to -\infty} \frac{3x+5}{\sqrt{9x^2} \sqrt{1 + \frac{2}{9x} + \frac{6}{9x^2}}} = \lim_{x \to -\infty} \frac{3x+5}{3|x| \sqrt{1 + \frac{2}{9x} + \frac{6}{9x^2}}}.$$\n\n5. **Since $x \to -\infty$, $|x| = -x$ (because $x$ is negative), so:**\n$$= \lim_{x \to -\infty} \frac{3x+5}{3(-x) \sqrt{1 + \frac{2}{9x} + \frac{6}{9x^2}}} = \lim_{x \to -\infty} \frac{3x+5}{-3x \sqrt{1 + \frac{2}{9x} + \frac{6}{9x^2}}}.$$\n\n6. **Divide numerator and denominator by $x$ to simplify:**\n$$= \lim_{x \to -\infty} \frac{\cancel{x}(3 + \frac{5}{x})}{\cancel{x}(-3) \sqrt{1 + \frac{2}{9x} + \frac{6}{9x^2}}} = \lim_{x \to -\infty} \frac{3 + \frac{5}{x}}{-3 \sqrt{1 + \frac{2}{9x} + \frac{6}{9x^2}}}.$$\n\n7. **Evaluate the limit by substituting $x \to -\infty$:**\nTerms with $\frac{1}{x}$ and $\frac{1}{x^2}$ go to zero, so:\n$$= \frac{3 + 0}{-3 \sqrt{1 + 0 + 0}} = \frac{3}{-3 \cdot 1} = -1.$$\n\n**Final answer:** $$\boxed{-1}.$$