1. **State the problem:** Simplify the expression $$\frac{(x + \sqrt{3})^2}{\sqrt{2} x^2 - 3\sqrt{2}} \cdot \frac{\sqrt{2}x - \sqrt{6}}{3}$$. 2. **Expand the numerator of the first fraction:** Use the formula $$(a+b)^2 = a^2 + 2ab + b^2$$ with $a = x$ and $b = \sqrt{3}$: $$ (x + \sqrt{3})^2 = x^2 + 2x\sqrt{3} + 3 $$ 3. **Rewrite the expression:** $$ \frac{x^2 + 2x\sqrt{3} + 3}{\sqrt{2} x^2 - 3\sqrt{2}} \cdot \frac{\sqrt{2}x - \sqrt{6}}{3} $$ 4. **Factor the denominator of the first fraction:** $$ \sqrt{2} x^2 - 3\sqrt{2} = \sqrt{2}(x^2 - 3) $$ 5. **Rewrite the expression with factored denominator:** $$ \frac{x^2 + 2x\sqrt{3} + 3}{\sqrt{2}(x^2 - 3)} \cdot \frac{\sqrt{2}x - \sqrt{6}}{3} $$ 6. **Notice that $x^2 - 3$ can be factored as $(x - \sqrt{3})(x + \sqrt{3})$:** $$ \frac{x^2 + 2x\sqrt{3} + 3}{\sqrt{2}(x - \sqrt{3})(x + \sqrt{3})} \cdot \frac{\sqrt{2}x - \sqrt{6}}{3} $$ 7. **Factor the numerator $x^2 + 2x\sqrt{3} + 3$ as a perfect square:** $$ (x + \sqrt{3})^2 $$ 8. **Rewrite the expression:** $$ \frac{(x + \sqrt{3})^2}{\sqrt{2}(x - \sqrt{3})(x + \sqrt{3})} \cdot \frac{\sqrt{2}x - \sqrt{6}}{3} $$ 9. **Factor the numerator of the second fraction:** $$ \sqrt{2}x - \sqrt{6} = \sqrt{2}(x - \sqrt{3}) $$ 10. **Rewrite the expression:** $$ \frac{(x + \sqrt{3})^2}{\sqrt{2}(x - \sqrt{3})(x + \sqrt{3})} \cdot \frac{\sqrt{2}(x - \sqrt{3})}{3} $$ 11. **Multiply the fractions:** $$ \frac{(x + \sqrt{3})^2 \cdot \sqrt{2}(x - \sqrt{3})}{\sqrt{2}(x - \sqrt{3})(x + \sqrt{3}) \cdot 3} $$ 12. **Cancel common factors $\sqrt{2}$ and $(x - \sqrt{3})$ in numerator and denominator:** $$ \frac{(x + \sqrt{3})^2 \cdot \cancel{\sqrt{2}} \cdot \cancel{(x - \sqrt{3})}}{\cancel{\sqrt{2}} \cdot \cancel{(x - \sqrt{3})} (x + \sqrt{3}) \cdot 3} = \frac{(x + \sqrt{3})^2}{(x + \sqrt{3}) \cdot 3} $$ 13. **Simplify the fraction by canceling one $(x + \sqrt{3})$ factor:** $$ \frac{\cancel{(x + \sqrt{3})} (x + \sqrt{3})}{\cancel{(x + \sqrt{3})} \cdot 3} = \frac{x + \sqrt{3}}{3} $$ **Final answer:** $$ \boxed{\frac{x + \sqrt{3}}{3}} $$