It is hard to see how they can not be linearly dependant since depending on the value of $x$, $f(x)$ is either equal to $g(x)$ or $-g(x)$. Also, their wronskian ends up being zero as well.
I can see that $g(x)$ is not differentiable at $x=0$, but I am not sure if it is enough to disqualify it from being linearly dependent. And even if they are not linearly dependent, would it then be right to call them linearly dependent?
Assume that $\alpha f+\beta g=0$ on $[-1,1]$, for some scalars $\alpha$ and $\beta$, Then, we have
$\alpha f(x)+\beta g(x)=0,~x \in [-1,1]$, implies $\alpha x^3+\beta x^2|x|=0$ for all $x \in [-1,1]$. In particular, $\alpha x^3+\beta x^2|x|=0$ for all $x \in [-1,0]$, gives $\alpha x^3-\beta x^3=0$ and thereby $\alpha-\beta=0$. Similarly, $\alpha x^3+\beta x^2|x|=0$ for all $x \in [0,1]$, gives $\alpha x^3+\beta x^3=0$ and thereby $\alpha+\beta=0$. Finally, $\alpha=\beta=0$.