Multiply $ \dfrac{3x-x^2}{3x^3} $ by $ \dfrac{x}{9-x^2} $ to get $ \dfrac{ -x^2 }{ -3x^4-9x^3 } $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{3x-x^2}{3x^3} \cdot \frac{x}{9-x^2} & \xlongequal{\text{Step 1}} \frac{ \left( -x \right) \cdot \color{blue}{ \left( x-3 \right) } }{ 3x^3 } \cdot \frac{ x }{ \left( -x-3 \right) \cdot \color{blue}{ \left( x-3 \right) } } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ -x }{ 3x^3 } \cdot \frac{ x }{ -x-3 } \xlongequal{\text{Step 3}} \frac{ \left( -x \right) \cdot x }{ 3x^3 \cdot \left( -x-3 \right) } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ -x^2 }{ -3x^4-9x^3 } \end{aligned} $$