Multiply $9$ by $ \dfrac{x^2}{2x+1} $ to get $ \dfrac{ 9x^2 }{ 2x+1 } $.

Step 1: Write $ 9 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 9 \cdot \frac{x^2}{2x+1} & \xlongequal{\text{Step 1}} \frac{9}{\color{red}{1}} \cdot \frac{x^2}{2x+1} \xlongequal{\text{Step 2}} \frac{ 9 \cdot x^2 }{ 1 \cdot \left( 2x+1 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 9x^2 }{ 2x+1 } \end{aligned} $$

Add $ \dfrac{4x^2-1}{3x} $ and $ \dfrac{9x^2}{2x+1} $ to get $ \dfrac{ \color{purple}{ 35x^3+4x^2-2x-1 } }{ 6x^2+3x }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ 2x+1 }$ and the second by $\color{blue}{ 3x }$.

$$ \begin{aligned} \frac{4x^2-1}{3x} + \frac{9x^2}{2x+1} & = \frac{ \left( 4x^2-1 \right) \cdot \color{blue}{ \left( 2x+1 \right) }}{ 3x \cdot \color{blue}{ \left( 2x+1 \right) }} + \frac{ 9x^2 \cdot \color{blue}{ 3x }}{ \left( 2x+1 \right) \cdot \color{blue}{ 3x }} = \\[1ex] &=\frac{ \color{purple}{ 8x^3+4x^2-2x-1 } }{ 6x^2+3x } + \frac{ \color{purple}{ 27x^3 } }{ 6x^2+3x }=\frac{ \color{purple}{ 35x^3+4x^2-2x-1 } }{ 6x^2+3x } \end{aligned} $$