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$$- \frac{2 \left(\left(3 - \frac{\left(3 x^{2} - 16\right)^{2}}{x^{2} \left(x^{2} - 16\right)}\right) \left(x - 1\right) + \frac{3 x^{2} - 16}{x}\right)}{x \left(x^{2} - 16\right)^{2}} = 0$$
Solve this equationThe roots of this equation
$$x_{1} = 2 \operatorname{CRootOf} {\left(3 x^{5} - 3 x^{4} + 4 x^{3} + 6 x^{2} - 8, 0\right)}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -4$$
$$x_{2} = 0$$
$$x_{3} = 4$$
$$\lim_{x \to -4^-}\left(- \frac{2 \left(\left(3 - \frac{\left(3 x^{2} - 16\right)^{2}}{x^{2} \left(x^{2} - 16\right)}\right) \left(x - 1\right) + \frac{3 x^{2} - 16}{x}\right)}{x \left(x^{2} - 16\right)^{2}}\right) = \infty$$
Let's take the limit$$\lim_{x \to -4^+}\left(- \frac{2 \left(\left(3 - \frac{\left(3 x^{2} - 16\right)^{2}}{x^{2} \left(x^{2} - 16\right)}\right) \left(x - 1\right) + \frac{3 x^{2} - 16}{x}\right)}{x \left(x^{2} - 16\right)^{2}}\right) = -\infty$$
Let's take the limit- the limits are not equal, so
$$x_{1} = -4$$
- is an inflection point
$$\lim_{x \to 0^-}\left(- \frac{2 \left(\left(3 - \frac{\left(3 x^{2} - 16\right)^{2}}{x^{2} \left(x^{2} - 16\right)}\right) \left(x - 1\right) + \frac{3 x^{2} - 16}{x}\right)}{x \left(x^{2} - 16\right)^{2}}\right) = -\infty$$
Let's take the limit$$\lim_{x \to 0^+}\left(- \frac{2 \left(\left(3 - \frac{\left(3 x^{2} - 16\right)^{2}}{x^{2} \left(x^{2} - 16\right)}\right) \left(x - 1\right) + \frac{3 x^{2} - 16}{x}\right)}{x \left(x^{2} - 16\right)^{2}}\right) = \infty$$
Let's take the limit- the limits are not equal, so
$$x_{2} = 0$$
- is an inflection point
$$\lim_{x \to 4^-}\left(- \frac{2 \left(\left(3 - \frac{\left(3 x^{2} - 16\right)^{2}}{x^{2} \left(x^{2} - 16\right)}\right) \left(x - 1\right) + \frac{3 x^{2} - 16}{x}\right)}{x \left(x^{2} - 16\right)^{2}}\right) = -\infty$$
Let's take the limit$$\lim_{x \to 4^+}\left(- \frac{2 \left(\left(3 - \frac{\left(3 x^{2} - 16\right)^{2}}{x^{2} \left(x^{2} - 16\right)}\right) \left(x - 1\right) + \frac{3 x^{2} - 16}{x}\right)}{x \left(x^{2} - 16\right)^{2}}\right) = \infty$$
Let's take the limit- the limits are not equal, so
$$x_{3} = 4$$
- 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
$$\left(-\infty, 2 \operatorname{CRootOf} {\left(3 x^{5} - 3 x^{4} + 4 x^{3} + 6 x^{2} - 8, 0\right)}\right]$$
Convex at the intervals
$$\left[2 \operatorname{CRootOf} {\left(3 x^{5} - 3 x^{4} + 4 x^{3} + 6 x^{2} - 8, 0\right)}, \infty\right)$$