1. Write the curves piecewise

Given: $y=x|x| \quad \text{and} \quad y=x-|x|$

We split according to the sign of $x$.

• For $x\ge 0$: $|x|=x$ so $y=x|x|=x^2, \qquad y=x-|x|=x-x=0$

• For $x<0$: $|x|=-x$ so $y=x|x|=x(-x)=-x^2, \qquad y=x-|x|=x-(-x)=2x$

Thus the curves are: $y=\begin{cases}x^2,&x\ge 0\\-x^2,&x<0\end{cases}$ and $y=\begin{cases}0,&x\ge 0\\2x,&x<0\end{cases}$

2. Find points of intersection

We solve separately.

For $x\ge 0$:

$x^2=0 \implies x=0$ So one intersection is: $(0,0)$

For $x<0$:

$-x^2=2x$ $x^2+2x=0$ $x(x+2)=0$ Since $x<0$, we take $x=-2$. Then $y=2(-2)=-4$ So another intersection is: $(-2,-4)$

Hence the bounded region lies between $x=-2$ and $x=0$.

3. Determine which curve is above the other on $[-2,0]$

On $x<0$, the curves are: $y_1=-x^2, \qquad y_2=2x$

Compare: $(-x^2)-(2x)=-(x^2+2x)=-x(x+2)$ For $-2<x<0$, this is positive, so $-x^2 > 2x$ Thus upper curve is $y=-x^2$ and lower curve is $y=2x$.

4. Compute the enclosed area

$A=\int_{-2}^{0}\big[(-x^2)-(2x)\big]dx$

$A=\int_{-2}^{0}(-x^2-2x)\,dx$

Integrate: $\int (-x^2-2x)dx=-\frac{x^3}{3}-x^2$

Now evaluate from $-2$ to $0$: $A=\left[-\frac{x^3}{3}-x^2\right]_{-2}^{0}$

At $x=0$: $0$

At $x=-2$: $-\frac{(-2)^3}{3}-(-2)^2=\frac{8}{3}-4=\frac{8-12}{3}=-\frac{4}{3}$

Therefore, $A=0-\left(-\frac{4}{3}\right)=\frac{4}{3}$

5. Check options

The enclosed area is $\boxed{\frac{4}{3}}$ So the correct option is C.