Multiply $5$ by $ \dfrac{x}{x^2-81} $ to get $ \dfrac{ 5x }{ x^2-81 } $.

Step 1: Write $ 5 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 5 \cdot \frac{x}{x^2-81} & \xlongequal{\text{Step 1}} \frac{5}{\color{red}{1}} \cdot \frac{x}{x^2-81} \xlongequal{\text{Step 2}} \frac{ 5 \cdot x }{ 1 \cdot \left( x^2-81 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 5x }{ x^2-81 } \end{aligned} $$

Subtract $ \dfrac{6}{x+9} $ from $ \dfrac{5x}{x^2-81} $ to get $ \dfrac{ \color{purple}{ -x+54 } }{ x^2-81 }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{x-9}$.

$$ \begin{aligned} \frac{5x}{x^2-81} - \frac{6}{x+9} & = \frac{ 5x }{ x^2-81 } - \frac{ 6 \cdot \color{blue}{ \left( x-9 \right) }}{ \left( x+9 \right) \cdot \color{blue}{ \left( x-9 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 5x } }{ x^2-81 } - \frac{ \color{purple}{ 6x-54 } }{ x^2-81 }=\frac{ \color{purple}{ 5x - \left( 6x-54 \right) } }{ x^2-81 } = \\[1ex] &=\frac{ \color{purple}{ -x+54 } }{ x^2-81 } \end{aligned} $$