Simple Absolute Value Inequality 
$$|\frac{x-1}{x+3}|\leq 2$$

I solve it as follow: 

$|\frac{x-1}{x+3}|\leq 2 \iff -2\leq \frac{x-1}{x+3} \leq 2 \iff -2\leq 1-\frac{4}{x+3} \leq 2 \iff -3\leq \frac{-4}{x+3} \leq 1 \iff \frac{3}{4}\geq \frac{1}{x+3} \geq -\frac{1}{4} \iff \frac{4}{3} \leq x+3 \leq -4 \iff -3+\frac{4}{3}=-\frac{5}{3}\leq x\leq -7$
I now see that I did not take into consideration $x\neq -3$ which can not be 
The book on the other side says that in answer is $x\leq -7$ or $x\geq -\frac{5}{3}$
Am I wrong?

Just break it into sections by sign:
$\left|\frac{x-1}{x+3}\right| =
\left\{
 \begin{array}{ll}
  \frac{x-1}{x+3}  & \mbox{if } x \geq 1 \\
  -\frac{x-1}{x+3}  & \mbox{if } x \in (-3,1) \\
  \frac{x-1}{x+3} & \mbox{if } x \lt -3
 \end{array}
\right.$
Then solve the three inequalities based on the sign of $x+3$:
$$x\geq1:\frac{x-1}{x+3} \leq 2 \iff x \geq -7$$
$$x\in(-3,1):-\frac{x-1}{x+3} \leq 2 \iff x \geq -\frac{5}{3}$$
$$x<-3:\frac{x-1}{x+3} \leq 2 \iff x \leq -7$$
That gives us the ranges $[1,\infty)$, $[-\frac53,1)$, and $(\infty,-7)$. So the final solution is $x \in (\infty,-7) \cup [-\frac53,\infty)$
