Multiply each term of $ \left( \color{blue}{y^2+8y+7}\right) $ by each term in $ \left( y^2+5y+6\right) $.

$$ \left( \color{blue}{y^2+8y+7}\right) \cdot \left( y^2+5y+6\right) = y^4+5y^3+6y^2+8y^3+40y^2+48y+7y^2+35y+42 $$

Combine like terms:

$$ y^4+ \color{blue}{5y^3} + \color{red}{6y^2} + \color{blue}{8y^3} + \color{green}{40y^2} + \color{orange}{48y} + \color{green}{7y^2} + \color{orange}{35y} +42 = \\ = y^4+ \color{blue}{13y^3} + \color{green}{53y^2} + \color{orange}{83y} +42 $$

Multiply each term of $ \left( \color{blue}{3y+9}\right) $ by each term in $ \left( y^2+y\right) $.

$$ \left( \color{blue}{3y+9}\right) \cdot \left( y^2+y\right) = 3y^3+3y^2+9y^2+9y $$

Simplify denominator

$$ 3y^3+ \color{blue}{3y^2} + \color{blue}{9y^2} +9y = 3y^3+ \color{blue}{12y^2} +9y $$

Simplify $ \dfrac{y^4+13y^3+53y^2+83y+42}{3y^3+12y^2+9y} $ to $ \dfrac{y^2+9y+14}{3y} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{y^2+4y+3}$.

$$ \begin{aligned} \frac{y^4+13y^3+53y^2+83y+42}{3y^3+12y^2+9y} & =\frac{ \left( y^2+9y+14 \right) \cdot \color{blue}{ \left( y^2+4y+3 \right) }}{ 3y \cdot \color{blue}{ \left( y^2+4y+3 \right) }} = \\[1ex] &= \frac{y^2+9y+14}{3y} \end{aligned} $$