Solving the absolute value inequality $\big| \frac{x}{x + 4} \big| < 4$ I was given this question and asked to find $x$:
$$\left| \frac{x}{x+4} \right|<4$$
I broke this into three pieces:
$$
\left| \frac{x}{x+4} \right| = \left\{
        \begin{array}{ll}
            \frac{x}{x+4} & \quad x > 0 \\
            -\frac{x}{x+4} & \quad -4 < x < 0 \\
            \frac{x}{x+4} & \quad x < −4
        \end{array}
    \right.
$$
Solving,
$$4>\frac{x}{x+4}$$
$$4x+16>x$$
$$3x>-16$$
$$x>-\frac{16}{3}=-5.3$$
and
$$4>-\frac{x}{x+4}$$
$$4x+16>-x$$
$$5x>-16$$
$$x>-3.2$$
The answer is $x<-5.3$ and $x>-3.2$. What am I doing wrong?

Your mistake was not considering the restrictions on $x$ when you solved the individual cases.
Case 1:  $x \geq 0$
If $x > 0$, then $x + 4 > 0$, so the direction of the inequality is preserved if we multiply by $x + 4$.  Hence,
\begin{align*}
\frac{x}{x + 4} & <  4\\
x & < 4(x + 4)\\
x & < 4x + 16\\
-16 & < 3x\\ 
-\frac{16}{3} & < x
\end{align*}
as you found.  However, the restriction that $x \geq 0$ means $x \geq 0$ and $x > -\dfrac{16}{3} \Longrightarrow x \geq 0$.  You did not account for the restriction $x \geq 0$.
Also, note that $-\dfrac{16}{3} \approx 5.3$.  The statement you made that $-\dfrac{16}{3} = 5.3$ is false since the decimal approximation you used does not actually equal the fraction.
Case 2:  $-4 < x < 0$
Since $-4 < x < 0$, $x + 4 > 0$.  Thus, the direction of the inequality is preserved if we multiply both sides of the inequality by $x + 4$.  Hence,
\begin{align*}
-\frac{x}{x + 4} & < 4\\
-x & < 4(x + 4)\\
-x & < 4x + 16\\
-16 & < 5x\\
-\frac{16}{5} & < x
\end{align*}
as you found.  The restriction that $-4 < x < 0$ means $-4 < x < 0$ and $-\dfrac{16}{5} < x \Longrightarrow -\dfrac{16}{5} < x < 0$.  You did not account for the restriction $x < 0$.
Case 3:  $x < -4$
If $x < -4$, then $x + 4 < 0$, so the direction of the inequality is reversed if we multiply both sides of the inequality by $x + 4$, which you did not take into account.
\begin{align*}
\frac{x}{x + 4} & < 4\\
x & > 4(x + 4)\\
x & > 4x + 16\\
-16 & > 3x\\
-\frac{16}{3} & > x
\end{align*}
The restriction that $x < -4$ means that $x < -4$ and $x < -\dfrac{16}{3} \Longrightarrow x < -\dfrac{16}{3}$.  
Combining the results of the three cases yields 
$$x \geq 0~\text{or}-\frac{16}{5} < x < 0~\text{or}~x < -\frac{16}{3} \Longrightarrow x > -\frac{16}{5}~\text{or}~x < -\frac{16}{3}$$
Hence, the solution set is 
$$S = \left(-\frac{16}{5}, \infty\right) \cup \left(-\infty, -\frac{16}{3}\right)$$
