Multiply $6$ by $ \dfrac{x}{5x-1} $ to get $ \dfrac{ 6x }{ 5x-1 } $.

Step 1: Write $ 6 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 6 \cdot \frac{x}{5x-1} & \xlongequal{\text{Step 1}} \frac{6}{\color{red}{1}} \cdot \frac{x}{5x-1} \xlongequal{\text{Step 2}} \frac{ 6 \cdot x }{ 1 \cdot \left( 5x-1 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 6x }{ 5x-1 } \end{aligned} $$

Subtract $ \dfrac{2}{x+3} $ from $ \dfrac{6x}{5x-1} $ to get $ \dfrac{ \color{purple}{ 6x^2+8x+2 } }{ 5x^2+14x-3 }$.

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}{ x+3 }$ and the second by $\color{blue}{ 5x-1 }$.

$$ \begin{aligned} \frac{6x}{5x-1} - \frac{2}{x+3} & = \frac{ 6x \cdot \color{blue}{ \left( x+3 \right) }}{ \left( 5x-1 \right) \cdot \color{blue}{ \left( x+3 \right) }} - \frac{ 2 \cdot \color{blue}{ \left( 5x-1 \right) }}{ \left( x+3 \right) \cdot \color{blue}{ \left( 5x-1 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 6x^2+18x } }{ 5x^2+15x-x-3 } - \frac{ \color{purple}{ 10x-2 } }{ 5x^2+15x-x-3 } = \\[1ex] &=\frac{ \color{purple}{ 6x^2+18x - \left( 10x-2 \right) } }{ 5x^2+14x-3 }=\frac{ \color{purple}{ 6x^2+8x+2 } }{ 5x^2+14x-3 } \end{aligned} $$