1) \( \frac{x}{3 x^{2} y}+\frac{x+1}{6 x y^{2}}-\frac{3}{9 x y} \)

To simplify the expression \( \frac{x}{3 x^{2} y}+\frac{x+1}{6 x y^{2}}-\frac{3}{9 x y} \), we first find a common denominator for all the fractions. The common denominator is \( 18 x^2 y^2 \). Now we rewrite each fraction with this common denominator: 1. For \( \frac{x}{3 x^{2} y} \): \[ \frac{x \cdot 6y}{3 x^{2} y \cdot 6y} = \frac{6xy}{18 x^{2} y^{2}} \] 2. For \( \frac{x+1}{6 x y^{2}} \): \[ \frac{(x+1) \cdot 3x^2}{6 x y^{2} \cdot 3x^2} = \frac{3x^2(x+1)}{18 x^{2} y^{2}} = \frac{3x^3 + 3x^2}{18 x^{2} y^{2}} \] 3. For \( -\frac{3}{9 x y} \): \[ -\frac{3 \cdot 2x y}{9 x y \cdot 2x y} = -\frac{6xy}{18 x^{2} y^{2}} \] Now we can combine these rewritten fractions: \[ \frac{6xy + 3x^3 + 3x^2 - 6xy}{18 x^{2} y^{2}} = \frac{3x^3 + 3x^2}{18 x^{2} y^{2}} \] Next, we can factor \( 3x^2 \) out in the numerator: \[ = \frac{3x^2(x + 1)}{18 x^{2} y^{2}} \] \[ = \frac{x + 1}{6y^{2}} \quad (x \neq 0) \] So, the simplified expression is: \[ \frac{x + 1}{6y^{2}} \quad (x \neq 0) \]