If a curve $\gamma$ is presented as a graph of a $C^2$-function $f$ then the standard $f''(x)=0$ condition produces the inflection points. But in the case at hand the given function, while continuous at $x=0$, is not differentiable there. Therefore we have to construct a regular parametric representation of the given graph $\gamma$ as a preliminary step.
Now the definition of $x^{1/3}$ is open to debate when $x<0$. In accordance with widespread practice I interpret this expression as follows:
$$x^{1/3}:={\rm sgn}(x)\>\root 3\of {|x|}\qquad(x\in{\mathbb R})\ .$$
It follows that $x(t):=t^3$ results in $x^{1/3}=t$, so that the graph $\gamma$ of $f$ can be written as
$$\gamma:\quad t\mapsto\cases{x(t)=t^3\cr y(t)=t^3-3t\cr}\qquad(-\infty<t<\infty)\ .\tag{1}$$
One computes
$$s'^2(t)=x'^2(t)+y'^2(t)=9(1-2t^2+2t^4)>0\qquad\forall t\ .$$
This shows that the representation $(1)$ of $\gamma$ is regular. The signed curvature $\kappa$ of $\gamma$ is given by
$$\kappa(t)={x'(t)y''(t)-x''(t)y'(t)\over s'^3(t)}={18 t\over s'^3(t)}\ .$$
This shows that $\kappa$ changes sign at the origin and implies that we have an inflection point of $\gamma$ at $(0,0)$. See the following figure:
