Subtract $ \dfrac{2}{y^2+6y+8} $ from $ \dfrac{3}{y^2+y-12} $ to get $ \dfrac{ \color{purple}{ y+12 } }{ y^3+3y^2-10y-24 }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ y+2 }$ and the second by $\color{blue}{ y-3 }$.

$$ \begin{aligned} \frac{3}{y^2+y-12} - \frac{2}{y^2+6y+8} & = \frac{ 3 \cdot \color{blue}{ \left( y+2 \right) }}{ \left( y^2+y-12 \right) \cdot \color{blue}{ \left( y+2 \right) }} - \frac{ 2 \cdot \color{blue}{ \left( y-3 \right) }}{ \left( y^2+6y+8 \right) \cdot \color{blue}{ \left( y-3 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 3y+6 } }{ y^3+2y^2+y^2+2y-12y-24 } - \frac{ \color{purple}{ 2y-6 } }{ y^3+2y^2+y^2+2y-12y-24 } = \\[1ex] &=\frac{ \color{purple}{ 3y+6 - \left( 2y-6 \right) } }{ y^3+3y^2-10y-24 }=\frac{ \color{purple}{ y+12 } }{ y^3+3y^2-10y-24 } \end{aligned} $$