Let's find the inflection points, we'll need to solve the equation for this
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative$$\frac{6 x^{2} \left(\frac{3 x^{3}}{x^{3} - 1} - 2\right)}{\left(x^{3} - 1\right)^{2}} = 0$$
Solve this equationThe roots of this equation
$$x_{1} = 0$$
$$x_{2} = - \sqrt[3]{2}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 1$$
$$\lim_{x \to 1^-}\left(\frac{6 x^{2} \left(\frac{3 x^{3}}{x^{3} - 1} - 2\right)}{\left(x^{3} - 1\right)^{2}}\right) = -\infty$$
$$\lim_{x \to 1^+}\left(\frac{6 x^{2} \left(\frac{3 x^{3}}{x^{3} - 1} - 2\right)}{\left(x^{3} - 1\right)^{2}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = 1$$
- is an inflection point
Сonvexity and concavity intervals:Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Concave at the intervals
$$\left(-\infty, - \sqrt[3]{2}\right]$$
Convex at the intervals
$$\left[- \sqrt[3]{2}, \infty\right)$$