Use the distributive property.

Subtract $ \dfrac{1}{x+1} $ from $ \dfrac{x}{x^2-1} $ to get $ \dfrac{ \color{purple}{ 1 } }{ x^2-1 }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{x-1}$.

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

Multiply $ \dfrac{1}{x^2-1} $ by $ 3 $ to get $ \dfrac{ 3 }{ x^2-1 } $.

Step 1: Write $ 3 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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