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$$\frac{2 + \frac{\left(6 x + \frac{1}{x^{2}}\right)^{2} \left(3 x^{2} - \frac{1}{x}\right)}{27 \left(1 - \frac{\left(3 x^{2} - \frac{1}{x}\right)^{2}}{9}\right)} - \frac{2}{3 x^{3}}}{\sqrt{1 - \frac{\left(3 x^{2} - \frac{1}{x}\right)^{2}}{9}}} = 0$$
Solve this equationThe roots of this equation
$$x_{1} = \frac{3^{\frac{2}{3}}}{3}$$
$$x_{2} = -0.303250105470311$$
$$x_{3} = 0.208358116435521$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(\frac{2 + \frac{\left(6 x + \frac{1}{x^{2}}\right)^{2} \left(3 x^{2} - \frac{1}{x}\right)}{27 \left(1 - \frac{\left(3 x^{2} - \frac{1}{x}\right)^{2}}{9}\right)} - \frac{2}{3 x^{3}}}{\sqrt{1 - \frac{\left(3 x^{2} - \frac{1}{x}\right)^{2}}{9}}}\right) = - \infty i$$
$$\lim_{x \to 0^+}\left(\frac{2 + \frac{\left(6 x + \frac{1}{x^{2}}\right)^{2} \left(3 x^{2} - \frac{1}{x}\right)}{27 \left(1 - \frac{\left(3 x^{2} - \frac{1}{x}\right)^{2}}{9}\right)} - \frac{2}{3 x^{3}}}{\sqrt{1 - \frac{\left(3 x^{2} - \frac{1}{x}\right)^{2}}{9}}}\right) = \infty i$$
- the limits are not equal, so
$$x_{1} = 0$$
- 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
$$\left[\frac{3^{\frac{2}{3}}}{3}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, \frac{3^{\frac{2}{3}}}{3}\right]$$