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

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{x+3}$.

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