$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{|x| + 3,}&{ifx \le - 3}\\{ - 2x,}&{if\; - 3 < x < 3}\\{6x + 2,}&{if\,x \ge 3}\end{array}} \right.$ }

At x $=$ 3 :

$\mathop {\lim }\limits_{x \to - {3^ - }} f(x) = \mathop {\lim }\limits_{x \to - {3^ - }} |x| + 3 = \mathop {\lim }\limits_{\scriptstyle x \to - 3 - h\atop\scriptstyle h \to 0} (| - 3 - h| + 3)$
$= | - 3 - 0| + 3 = 3 + 3 = 6$

$\mathop {\lim }\limits_{x \to - {3^ + }} f(x) = \mathop {\lim }\limits_{x \to - {3^ + }} ( - 2x) = \mathop {\lim }\limits_{\scriptstyle x \to - 3 + h\atop\scriptstyle h \to 0} ( - 2( - 3 + h3))$

$= - 2( - 3 + 0) = 6$
$f( - 3) = | - 3| + 3 = 3 + 3 = 6$

Thus,, $\mathop {\lim }\limits_{x \to - {3^ - }} f(x) = \mathop {\lim }\limits_{x \to - {3^ + }} f(x) = f( - 3)$

therefore, f is continuous at $x = - 3.$

At x $=$ 3 :

$\mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{x \to {3^ - }} ( - 2x) = \mathop {\lim }\limits_{\scriptstyle x \to 3 - h\atop\scriptstyle h \to 0} ( - 2(3 - h)) = - 2(3 - 0) = - 6$

$\mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{x \to {3^ + }} (6x + 2) = \mathop {\lim }\limits_{\scriptstyle x \to 3 + h\atop\scriptstyle h \to 0} (6(3 + h) + 2)$
$= 6(3 + 0) + 2 = 20$

Thus, $\mathop {\lim }\limits_{x \to {3^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {3^ + }} f(x)$

Hence we can say that $f(x)$ is discontinuous at x $=$ 3.

So, the only point of discontinuity of the given function f is 3.