Subtract $ \dfrac{2}{3x} $ from $ \dfrac{3}{3x^2-9x} $ to get $ \dfrac{ \color{purple}{ -6x^2+27x } }{ 9x^3-27x^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}{ 3x }$ and the second by $\color{blue}{ 3x^2-9x }$.

$$ \begin{aligned} \frac{3}{3x^2-9x} - \frac{2}{3x} & = \frac{ 3 \cdot \color{blue}{ 3x }}{ \left( 3x^2-9x \right) \cdot \color{blue}{ 3x }} - \frac{ 2 \cdot \color{blue}{ \left( 3x^2-9x \right) }}{ 3x \cdot \color{blue}{ \left( 3x^2-9x \right) }} = \\[1ex] &=\frac{ \color{purple}{ 9x } }{ 9x^3-27x^2 } - \frac{ \color{purple}{ 6x^2-18x } }{ 9x^3-27x^2 }=\frac{ \color{purple}{ 9x - \left( 6x^2-18x \right) } }{ 9x^3-27x^2 } = \\[1ex] &=\frac{ \color{purple}{ -6x^2+27x } }{ 9x^3-27x^2 } \end{aligned} $$