The function $\mathit{\text{f}}\left(x\right)=2\left|x\right|+\left|x+2\right|-\left|\left|x+2\right|-2\left|x\right|\right|$ has a local minimum or a local maximum at $x=$

$f\left(x\right)=2\left|x\right|+\left|x+2\right|-\left|\left|x+2\right|-2\left|x\right|\right|$

CASE 1:

For $x\le -2,$

$\left|x\right|=-x$ and $\left|x+2\right|=-\left(x+2\right)$ $(\because x, x+2<0)$

$\Rightarrow f\left(x\right)=-3x-2-\left|x-2\right|$

$=-3x-2+(x-2) (\because x-2<0)$

$=-2x-4 ...\left(1\right)$

CASE 2:

For $-2<x\le 0,$

$\left|x\right|=-x \left(\because x<0\right) \& \left|x+2\right|=(x+2) (\because x+2>0)$

$\Rightarrow f\left(x\right)=-2x+(x+2)-\left|(x+2)+2x\right|=-x+2-\left|3x+2\right|$

$=\left\{\begin{array}{c}-x+2+(3x+2), -2<x\le \frac{-2}{3}\\ -x+2-(3x+2), \frac{-2}{3}<x\le 0\end{array}\right\}$

$=\left\{\begin{array}{c}2x+4, -2<x\le \frac{-2}{3}\\ -4x, \frac{-2}{3}<x\le 0\end{array}\right\} ...\left(2\right)$

CASE 3:

For $x>0,$

$\left|x\right|=x \& \left|x+2\right|=(x+2)$

$\Rightarrow f\left(x\right)=2x+(x+2)-\left|(x+2)-2x\right|=3x+2-\left|-x+2\right|$

$=\left\{\begin{array}{c}3x+2-(-x+2), 0<x\le 2\\ 3x+2+(-x+2), 2<x \end{array}\right\}$

$=\left\{\begin{array}{c}4x, 0<x\le 2\\ 2x+4, 2<x\end{array}\right\} ...\left(3\right)$

From $\left(1\right), \left(2\right) \& \left(3\right),$ we get
$\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{)=}\left\{\begin{array}{cc}-2x-4,& x\le -2\\ 2x+4,& -2<x\le \frac{-2}{3}\\ -4x,& \frac{-2}{3}<x\le 0\\ 4x,& 0<x\le 2\\ 2x+4,& x>2\end{array}\right\}$

From graph, $y=f\left(x\right)$ has local minima at $x=-2, 0$ and local maxima at $x=\frac{-2}{3}.$