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$$\frac{2 \left(- \frac{6 x^{3}}{x^{3} + 1} + \frac{3 x \left(x^{2} + 1\right) \left(\frac{3 x^{3}}{x^{3} + 1} - 1\right)}{x^{3} + 1} + 1\right)}{x^{3} + 1} = 0$$
Solve this equationThe roots of this equation
$$x_{1} = 0.290553628621719$$
$$x_{2} = 1.16555408630366$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -1$$
$$\lim_{x \to -1^-}\left(\frac{2 \left(- \frac{6 x^{3}}{x^{3} + 1} + \frac{3 x \left(x^{2} + 1\right) \left(\frac{3 x^{3}}{x^{3} + 1} - 1\right)}{x^{3} + 1} + 1\right)}{x^{3} + 1}\right) = -\infty$$
$$\lim_{x \to -1^+}\left(\frac{2 \left(- \frac{6 x^{3}}{x^{3} + 1} + \frac{3 x \left(x^{2} + 1\right) \left(\frac{3 x^{3}}{x^{3} + 1} - 1\right)}{x^{3} + 1} + 1\right)}{x^{3} + 1}\right) = \infty$$
- 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
$$\left(-\infty, 0.290553628621719\right] \cup \left[1.16555408630366, \infty\right)$$
Convex at the intervals
$$\left[0.290553628621719, 1.16555408630366\right]$$