1. **State the problem:** Solve the inequality $$6^x - 9 \cdot 2^x + 3^{x+2} - 9^x > 0.$$\n\n2. **Rewrite terms to express bases consistently:** Note that $$6^x = (2 \cdot 3)^x = 2^x \cdot 3^x$$ and $$9 = 3^2,$$ so $$9 \cdot 2^x = 3^2 \cdot 2^x,$$ and $$9^x = (3^2)^x = 3^{2x}.$$ Also, $$3^{x+2} = 3^x \cdot 3^2 = 9 \cdot 3^x.$$\n\n3. **Rewrite the inequality:**\n$$6^x - 9 \cdot 2^x + 3^{x+2} - 9^x > 0$$\nbecomes\n$$2^x 3^x - 9 2^x + 9 3^x - 3^{2x} > 0.$$\n\n4. **Group terms:**\nGroup terms with $$2^x$$ and $$3^x$$:\n$$2^x 3^x - 9 2^x + 9 3^x - 3^{2x} = 2^x (3^x - 9) + 3^x (9 - 3^x).$$\n\n5. **Substitute variables:** Let $$a = 2^x > 0$$ and $$b = 3^x > 0.$$ The inequality becomes\n$$a(b - 9) + b(9 - b) > 0.$$\n\n6. **Simplify:**\n$$a b - 9 a + 9 b - b^2 > 0.$$\nRewrite as\n$$a b - 9 a + 9 b - b^2 > 0.$$\n\n7. **Group terms:**\n$$a b - 9 a + 9 b - b^2 = a b - 9 a + 9 b - b^2 = a(b - 9) + b(9 - b).$$\n\n8. **Rewrite inequality:**\n$$a(b - 9) + b(9 - b) > 0.$$\n\n9. **Analyze the inequality:**\nRewrite as\n$$a(b - 9) > b(b - 9).$$\n\n10. **Consider cases based on $$b - 9$$:**\n- If $$b - 9 > 0$$ (i.e., $$b > 9$$), divide both sides by $$b - 9$$ (positive), inequality direction stays the same:\n$$a > b.$$\n- If $$b - 9 < 0$$ (i.e., $$b < 9$$), divide both sides by $$b - 9$$ (negative), inequality direction reverses:\n$$a < b.$$\n- If $$b = 9$$, then the inequality becomes $$0 > 0$$ which is false.\n\n11. **Recall $$a = 2^x$$ and $$b = 3^x$$:**\n- When $$b > 9$$, i.e., $$3^x > 9 = 3^2$$, so $$x > 2$$, inequality is $$2^x > 3^x$$.\n- When $$b < 9$$, i.e., $$x < 2$$, inequality is $$2^x < 3^x$$.\n\n12. **Analyze $$2^x$$ and $$3^x$$:**\n- For $$x > 0$$, $$3^x > 2^x$$ because base 3 is larger than base 2.\n- For $$x < 0$$, $$2^x > 3^x$$ because negative exponents invert the bases and $$\frac{1}{2^|x|} > \frac{1}{3^{|x|}}.$$\n\n13. **Check the two cases:**\n- For $$x > 2$$, inequality requires $$2^x > 3^x$$, which is false since $$3^x > 2^x$$ for positive $$x$$.\n- For $$x < 2$$, inequality requires $$2^x < 3^x$$, which is true for all $$x > 0$$ but false for $$x < 0$$.\n\n14. **Check the boundary at $$x=2$$:**\nAt $$x=2$$, original expression is zero, so not included.\n\n15. **Check sign for $$x < 0$$:**\nTry $$x = -1$$:\n$$6^{-1} - 9 \cdot 2^{-1} + 3^{1} - 9^{-1} = \frac{1}{6} - 9 \cdot \frac{1}{2} + 3 - \frac{1}{9} = \frac{1}{6} - \frac{9}{2} + 3 - \frac{1}{9}.$$\nCalculate:\n$$\frac{1}{6} \approx 0.1667, \quad -\frac{9}{2} = -4.5, \quad 3, \quad -\frac{1}{9} \approx -0.1111.$$\nSum:\n$$0.1667 - 4.5 + 3 - 0.1111 = -1.4444 < 0,$$ so inequality is false for $$x = -1$$.\n\n16. **Check sign for $$0 < x < 2$$:**\nTry $$x=1$$:\n$$6^1 - 9 \cdot 2^1 + 3^{3} - 9^1 = 6 - 18 + 27 - 9 = 6 - 18 + 27 - 9 = 6.$$\nPositive, so inequality holds for $$0 < x < 2$$.\n\n17. **Check sign for $$x > 2$$:**\nTry $$x=3$$:\n$$6^3 - 9 \cdot 2^3 + 3^{5} - 9^3 = 216 - 72 + 243 - 729 = 216 - 72 + 243 - 729 = -342.$$\nNegative, inequality false for $$x > 2$$.\n\n18. **Check sign for $$x=0$$:**\n$$6^0 - 9 \cdot 2^0 + 3^{2} - 9^0 = 1 - 9 + 9 - 1 = 0,$$ not greater than zero.\n\n19. **Summary:**\nInequality holds for $$x$$ in the interval $$(0, 2).$$\n\n**Final answer:** $$\boxed{0 < x < 2}.$$