1. **State the problem:** Simplify the expression $$\left( \frac{1}{x-4} - \frac{1}{x+4\sqrt{x}+4} \right) : \frac{\sqrt{x}}{x+2\sqrt{x}}.$$\n\n2. **Rewrite the division as multiplication by the reciprocal:**\n$$\left( \frac{1}{x-4} - \frac{1}{x+4\sqrt{x}+4} \right) \times \frac{x+2\sqrt{x}}{\sqrt{x}}.$$\n\n3. **Factor expressions where possible:**\nNote that $$x+4\sqrt{x}+4 = (\sqrt{x}+2)^2,$$ and $$x-4 = (\sqrt{x})^2 - 2^2 = (\sqrt{x}-2)(\sqrt{x}+2).$$ Also, $$x+2\sqrt{x} = \sqrt{x}(\sqrt{x}+2).$$\n\n4. **Rewrite the expression using these factorizations:**\n$$\left( \frac{1}{(\sqrt{x}-2)(\sqrt{x}+2)} - \frac{1}{(\sqrt{x}+2)^2} \right) \times \frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}}.$$\n\n5. **Simplify the multiplication factor:**\n$$\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}} = \cancel{\frac{\sqrt{x}}{\sqrt{x}}}(\sqrt{x}+2) = \sqrt{x}+2.$$\n\n6. **Find common denominator for the subtraction inside parentheses:**\nThe denominators are $(\sqrt{x}-2)(\sqrt{x}+2)$ and $(\sqrt{x}+2)^2$. The common denominator is $(\sqrt{x}+2)^2(\sqrt{x}-2)$.\n\nRewrite each fraction:\n$$\frac{1}{(\sqrt{x}-2)(\sqrt{x}+2)} = \frac{\sqrt{x}+2}{(\sqrt{x}+2)^2(\sqrt{x}-2)},$$\n$$\frac{1}{(\sqrt{x}+2)^2} = \frac{\sqrt{x}-2}{(\sqrt{x}+2)^2(\sqrt{x}-2)}.$$\n\n7. **Subtract the fractions:**\n$$\frac{\sqrt{x}+2}{(\sqrt{x}+2)^2(\sqrt{x}-2)} - \frac{\sqrt{x}-2}{(\sqrt{x}+2)^2(\sqrt{x}-2)} = \frac{(\sqrt{x}+2) - (\sqrt{x}-2)}{(\sqrt{x}+2)^2(\sqrt{x}-2)} = \frac{4}{(\sqrt{x}+2)^2(\sqrt{x}-2)}.$$\n\n8. **Multiply by the factor outside the parentheses:**\n$$\frac{4}{(\sqrt{x}+2)^2(\sqrt{x}-2)} \times (\sqrt{x}+2) = \frac{4(\sqrt{x}+2)}{(\sqrt{x}+2)^2(\sqrt{x}-2)} = \frac{4}{\cancel{(\sqrt{x}+2)}(\sqrt{x}-2)}.$$\n\n9. **Cancel common factor:**\n$$\frac{4}{\cancel{(\sqrt{x}+2)}(\sqrt{x}-2)} = \frac{4}{(\sqrt{x}-2)(\sqrt{x}+2)} = \frac{4}{x-4}.$$\n\n**Final answer:** $$\boxed{\frac{4}{x-4}}.$$