Multiply $ \dfrac{x-3}{xy^2} $ by $ \dfrac{4x^2y^3}{x^2+x-12} $ to get $ \dfrac{ 4x^2y^3 }{ x^2y^2+4xy^2 } $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{x-3}{xy^2} \cdot \frac{4x^2y^3}{x^2+x-12} & \xlongequal{\text{Step 1}} \frac{ 1 \cdot \color{blue}{ \left( x-3 \right) } }{ xy^2 } \cdot \frac{ 4x^2y^3 }{ \left( x+4 \right) \cdot \color{blue}{ \left( x-3 \right) } } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 1 }{ xy^2 } \cdot \frac{ 4x^2y^3 }{ x+4 } \xlongequal{\text{Step 3}} \frac{ 1 \cdot 4x^2y^3 }{ xy^2 \cdot \left( x+4 \right) } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ 4x^2y^3 }{ x^2y^2+4xy^2 } \end{aligned} $$