Subtract $ \dfrac{3}{x+2} $ from $ \dfrac{x}{x+1} $ to get $ \dfrac{ \color{purple}{ x^2-x-3 } }{ x^2+3x+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}{ x+2 }$ and the second by $\color{blue}{ x+1 }$.

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