Subtract $ \dfrac{2}{t-3} $ from $ \dfrac{2}{t+3} $ to get $ \dfrac{ \color{purple}{ -12 } }{ t^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}{ t-3 }$ and the second by $\color{blue}{ t+3 }$.

$$ \begin{aligned} \frac{2}{t+3} - \frac{2}{t-3} & = \frac{ 2 \cdot \color{blue}{ \left( t-3 \right) }}{ \left( t+3 \right) \cdot \color{blue}{ \left( t-3 \right) }} - \frac{ 2 \cdot \color{blue}{ \left( t+3 \right) }}{ \left( t-3 \right) \cdot \color{blue}{ \left( t+3 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 2t-6 } }{ t^2 -\cancel{3t}+ \cancel{3t}-9 } - \frac{ \color{purple}{ 2t+6 } }{ t^2 -\cancel{3t}+ \cancel{3t}-9 } = \\[1ex] &=\frac{ \color{purple}{ 2t-6 - \left( 2t+6 \right) } }{ t^2-9 }=\frac{ \color{purple}{ -12 } }{ t^2-9 } \end{aligned} $$