1. The problem is to understand how the factor $\frac{1}{9}$ is taken out in the expression $$\frac{1}{3} x e^{3x} - \frac{1}{9} e^{3x} + C = \frac{1}{9} (3x - 1) e^{3x} + C.$$\n\n2. Start with the expression before factoring: $$\frac{1}{3} x e^{3x} - \frac{1}{9} e^{3x}.$$\n\n3. Notice that both terms contain $e^{3x}$, so we can factor $e^{3x}$ out: $$e^{3x} \left( \frac{1}{3} x - \frac{1}{9} \right).$$\n\n4. Now, look inside the parentheses: $$\frac{1}{3} x - \frac{1}{9}.$$\n\n5. To factor out $\frac{1}{9}$ from these terms, write each term with denominator 9: $$\frac{3}{9} x - \frac{1}{9} = \frac{1}{9} (3x - 1).$$\n\n6. So the entire expression becomes: $$e^{3x} \cdot \frac{1}{9} (3x - 1) = \frac{1}{9} (3x - 1) e^{3x}.$$\n\n7. This shows how the $\frac{1}{9}$ is factored out by expressing $\frac{1}{3} x$ as $\frac{3}{9} x$ and factoring the common denominator $\frac{1}{9}$ from both terms inside the parentheses.