Subtract $ \dfrac{1}{t+3i} $ from $ \dfrac{1}{t+i} $ to get $ \dfrac{ \color{purple}{ 2i } }{ 3i^2+4it+t^2 }$.

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

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