Subtract $ \dfrac{2}{x-3} $ from $ \dfrac{x}{x+3} $ to get $ \dfrac{ \color{purple}{ x^2-5x-6 } }{ x^2-9 }$.

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-3 }$ and the second by $\color{blue}{ x+3 }$.

$$ \begin{aligned} \frac{x}{x+3} - \frac{2}{x-3} & = \frac{ x \cdot \color{blue}{ \left( x-3 \right) }}{ \left( x+3 \right) \cdot \color{blue}{ \left( x-3 \right) }} - \frac{ 2 \cdot \color{blue}{ \left( x+3 \right) }}{ \left( x-3 \right) \cdot \color{blue}{ \left( x+3 \right) }} = \\[1ex] &=\frac{ \color{purple}{ x^2-3x } }{ x^2 -\cancel{3x}+ \cancel{3x}-9 } - \frac{ \color{purple}{ 2x+6 } }{ x^2 -\cancel{3x}+ \cancel{3x}-9 } = \\[1ex] &=\frac{ \color{purple}{ x^2-3x - \left( 2x+6 \right) } }{ x^2-9 }=\frac{ \color{purple}{ x^2-5x-6 } }{ x^2-9 } \end{aligned} $$