Simplify denominator

$$ x^2 \, \color{blue}{ -\cancel{4}} \,+ \, \color{blue}{ \cancel{4}} \, = x^2 $$

Multiply $4$ by $ \dfrac{x}{x^2} $ to get $ \dfrac{ 4x }{ x^2 } $.

Step 1: Write $ 4 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 4 \cdot \frac{x}{x^2} & \xlongequal{\text{Step 1}} \frac{4}{\color{red}{1}} \cdot \frac{x}{x^2} \xlongequal{\text{Step 2}} \frac{ 4 \cdot x }{ 1 \cdot x^2 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 4x }{ x^2 } \end{aligned} $$

Multiply $3$ by $ \dfrac{x}{x^2+4} $ to get $ \dfrac{ 3x }{ x^2+4 } $.

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.

$$ \begin{aligned} 3 \cdot \frac{x}{x^2+4} & \xlongequal{\text{Step 1}} \frac{3}{\color{red}{1}} \cdot \frac{x}{x^2+4} \xlongequal{\text{Step 2}} \frac{ 3 \cdot x }{ 1 \cdot \left( x^2+4 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 3x }{ x^2+4 } \end{aligned} $$

Subtract $ \dfrac{3x}{x^2+4} $ from $ \dfrac{4x}{x^2} $ to get $ \dfrac{ \color{purple}{ x^3+16x } }{ x^4+4x^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+4 }$ and the second by $\color{blue}{ x^2 }$.

$$ \begin{aligned} \frac{4x}{x^2} - \frac{3x}{x^2+4} & = \frac{ 4x \cdot \color{blue}{ \left( x^2+4 \right) }}{ x^2 \cdot \color{blue}{ \left( x^2+4 \right) }} - \frac{ 3x \cdot \color{blue}{ x^2 }}{ \left( x^2+4 \right) \cdot \color{blue}{ x^2 }} = \\[1ex] &=\frac{ \color{purple}{ 4x^3+16x } }{ x^4+4x^2 } - \frac{ \color{purple}{ 3x^3 } }{ x^4+4x^2 }=\frac{ \color{purple}{ x^3+16x } }{ x^4+4x^2 } \end{aligned} $$