Multiply $2$ by $ \dfrac{x}{x^2+5x+6} $ to get $ \dfrac{ 2x }{ x^2+5x+6 } $.

Step 1: Write $ 2 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 2 \cdot \frac{x}{x^2+5x+6} & \xlongequal{\text{Step 1}} \frac{2}{\color{red}{1}} \cdot \frac{x}{x^2+5x+6} \xlongequal{\text{Step 2}} \frac{ 2 \cdot x }{ 1 \cdot \left( x^2+5x+6 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2x }{ x^2+5x+6 } \end{aligned} $$

Subtract $ \dfrac{x-3}{x^2+2x+3} $ from $ \dfrac{2x}{x^2+5x+6} $ to get $ \dfrac{ \color{purple}{ x^3+2x^2+15x+18 } }{ x^4+7x^3+19x^2+27x+18 }$.

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}{ x^2+2x+3 }$ and the second by $\color{blue}{ x^2+5x+6 }$.

$$ \begin{aligned} \frac{2x}{x^2+5x+6} - \frac{x-3}{x^2+2x+3} & = \frac{ 2x \cdot \color{blue}{ \left( x^2+2x+3 \right) }}{ \left( x^2+5x+6 \right) \cdot \color{blue}{ \left( x^2+2x+3 \right) }} - \frac{ \left( x-3 \right) \cdot \color{blue}{ \left( x^2+5x+6 \right) }}{ \left( x^2+2x+3 \right) \cdot \color{blue}{ \left( x^2+5x+6 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 2x^3+4x^2+6x } }{ x^4+2x^3+3x^2+5x^3+10x^2+15x+6x^2+12x+18 } - \frac{ \color{purple}{ x^3+5x^2+6x-3x^2-15x-18 } }{ x^4+2x^3+3x^2+5x^3+10x^2+15x+6x^2+12x+18 } = \\[1ex] &=\frac{ \color{purple}{ 2x^3+4x^2+6x - \left( x^3+5x^2+6x-3x^2-15x-18 \right) } }{ x^4+7x^3+19x^2+27x+18 } = \\[1ex] &=\frac{ \color{purple}{ x^3+2x^2+15x+18 } }{ x^4+7x^3+19x^2+27x+18 } \end{aligned} $$