Question: If a given first derivative is: $\ {dy \over dx} = {-48x \over (x^2+12)^2} $

What are the steps using the quotient rule to derive the second derivative: $\ {d^2y \over dx^2} = {-144(4-x^2) \over (x^2+12)^3} $

My Steps:

\begin{align\*}
{d^2y \over dx^2}
&= {-48(x^2 +12)^2 - 2(x^2+12)(2x)(-48x)\over (x^2+12)^4} \\
&=-48{(x^2 +12)^2 - 4x^2(x^2+12)\over (x^2+12)^4} \\
&= -48{(x^2 +12)( - 4x^2 +(x^2+12))\over (x^2+12)^4} \\
&= -48{(x^2 +12)( - 4x^2)\over (x^2+12)^3} \\
&= {???}
\end{align\*}

Answer: 
We can factor out the constant term to make life easier (and just multiply by that $-48$ at the end of our calculation) as:

$$-48 \dfrac{x}{(x^2+12)^2}$$

This makes it easier to use the quotient and chain rule (I will assume you know these).

The derivative of $\ {dy \over dx} = {-48x \over (x^2+12)^2} $, using the quotient and chain rule is:

$\dfrac{d^2y}{dx^2} = -48 \dfrac{(1)(x^2+12)^2 - 2(x^2+12)(2x)x}{(x^2+12)^4} = -48 \dfrac{(x^2+12) - 2(2x)x}{(x^2+12)^3} = \dfrac{-144(4-x^2)}{(x^2+12)^3}$
First, make sure you know what the quotient rule is - which you can derive from the product rule say by taking the derivative of $g(x)/f(x)$.

Then, simplify. Looking at the product rule, expect to have to cancel a factor of $x^2+12$ from the top and bottom.