Multiply $7$ by $ \dfrac{x}{x+6} $ to get $ \dfrac{ 7x }{ x+6 } $.

Step 1: Write $ 7 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 7 \cdot \frac{x}{x+6} & \xlongequal{\text{Step 1}} \frac{7}{\color{red}{1}} \cdot \frac{x}{x+6} \xlongequal{\text{Step 2}} \frac{ 7 \cdot x }{ 1 \cdot \left( x+6 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 7x }{ x+6 } \end{aligned} $$

Subtract $ \dfrac{7x}{x+6} $ from $ \dfrac{3}{x^2+9x+18} $ to get $ \dfrac{ \color{purple}{ -7x^2-21x+3 } }{ x^2+9x+18 }$.

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

$$ \begin{aligned} \frac{3}{x^2+9x+18} - \frac{7x}{x+6} & = \frac{ 3 }{ x^2+9x+18 } - \frac{ 7x \cdot \color{blue}{ \left( x+3 \right) }}{ \left( x+6 \right) \cdot \color{blue}{ \left( x+3 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 3 } }{ x^2+9x+18 } - \frac{ \color{purple}{ 7x^2+21x } }{ x^2+9x+18 }=\frac{ \color{purple}{ 3 - \left( 7x^2+21x \right) } }{ x^2+9x+18 } = \\[1ex] &=\frac{ \color{purple}{ -7x^2-21x+3 } }{ x^2+9x+18 } \end{aligned} $$