You have the following expression:
[tex]\frac{-4x^2+16}{(x^2+4)^2}[/tex]The previous expression can be written as follow:
[tex](-4x^2+16)\cdot\frac{1}{(x^2+4)^2}[/tex]The derivative of the previous expression is the derivative of a product:
[tex]\begin{gathered} (-4x^2+16)^{\prime}\cdot\frac{1}{(x^2+16)^2}+(-4x^2+16)\cdot(\frac{1}{(x^2+4)^2})^{\prime} \\ =(-8x)\cdot\frac{1}{(x^2+16)^2}+(-4x^2+16)\cdot(-2)(x^2+4)^{-3}\cdot(2x) \\ =\frac{-8x}{(x^2+16)^2}-\frac{4x(-4x^2+16)}{(x^2+4)^3} \end{gathered}[/tex]by factorizing the numerator of the second term, you obtain:
[tex]\begin{gathered} \frac{-8x}{(x^2+16)^2}+\frac{4x(x^2-16)}{(x^2+4)^3} \\ =\frac{-8x}{(x^2+16)^2}+\frac{4x(x+4)(x-4)}{(x^2+4)^3} \\ =\frac{-8x}{(x^2+16)^2}+\frac{4x(x-4)}{(x^2+4)^2} \end{gathered}[/tex]