Subtract $ \dfrac{4}{x-2} $ from $ \dfrac{x}{x+3} $ to get $ \dfrac{ \color{purple}{ x^2-6x-12 } }{ x^2+x-6 }$.

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

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