What is inflection point? $$f(x)=x-3x^{1/3}$$
$$f'(x)=1-x^{-2/3}$$
$$f''(x)=\frac{2}{3}x^{-5/3}$$
$f ''(x)=0$ has no solution. Does that mean f has no inflection point? If $x > 0$ then $f$ is concave up and if $x < 0$ then $f$ is concave down. Where $f$ changes concavity?
Inflection point is where the function changes concavity or where the second derivative is zero or ....?
 A: An inflection point occurs at points of the domain at which the function changes concavity. These need not be points at which the second derivative (or even the first) exists. Now, if your function's second derivative is defined and continuous in a neighborhood of an inflection point, then it will be zero at that point.
A: 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:

