The function $f\left(x\right)=\frac{9x}{x+2}$ if x< 1

$=3$ if  $x=1$

$=\frac{x+3}{2}$

 if x> 1 is continuous 

$\begin{array}{l}\underset{x\to 1^{-}}{Lt}f\left(x\right)=\underset{x\to 1^{-}}{Lt}\frac{9x}{x+2}=3\\ \underset{x\to 1^{+}}{Lt}f\left(x\right)=\underset{x\to 1^{-}}{Lt}\frac{x+3}{2}=2\\ \Rightarrow \underset{x\to 1^{-}}{Lt}f\left(x\right)\ne \underset{x\to 1^{+}}{Lt}f\left(x\right)\end{array}$

$\therefore f\left(x\right)$ is discontinuous at $x=1$ 

$ifx=-2;f\left(x\right)=\frac{9x}{x+2}$ is not defined

$f\left(x\right)$ is continuous on $\Re -\{-2,1\}$ is $(-\infty ,-2)\cup (-2,1)\cup (1,\infty )$