Given: 

$f\left(x\right)=\left\{\begin{array}{l}2+2x, x\in [-1,0)\\ 1-\frac{x}{3}, x\in \left[0,3\right]\end{array}\right.$ and $g\left(x\right)=\left\{\begin{array}{l}x, x\in (0,1]\\ -x, x\in \left[-3,0\right]\end{array}\right.$

$\Rightarrow f\left(g\left(x\right)\right)=\left\{\begin{array}{l}2+2g\left(x\right), g\left(x\right)\in [-1,0)\\ 1-\frac{g\left(x\right)}{3}, g\left(x\right)\in \left[0,3\right]\end{array}\right.$

$\Rightarrow f\left(g\left(x\right)\right)=\left\{\begin{array}{l}2+2\left|x\right|, \left|x\right|\in [-1,0)\Rightarrow x\in \varphi \\ 1-\frac{\left|x\right|}{3}, \left|x\right|\in \left[0,3\right]\end{array}\right.$

$\left\{\text{as} x\in \left[-3,3\right] \text{for }\left|x\right|\le 3 \text{but }x\in \left[-3,1\right] \text{in }g\left(x\right)\right\}$

$\Rightarrow f\left(g\left(x\right)\right)=1-\frac{\left|x\right|}{3}, x\in \left[-3,1\right]$

Now, maximum value of $f\left(g\left(x\right)\right)=1-\frac{\left|x\right|}{3}=1-\frac{0}{3}=1$ 

And minimum value is given by $f\left(g\left(x\right)\right)=1-\frac{3}{3}=1-1=0$

Hence, the range of the function will be $\left[0,1\right]$