For $x\ge -3\Rightarrow f''(x)=\dfrac{3(2x^2-x-1)}{(x^2+1)^{\frac{5}{2}}}$
$2x^2-x-1=0$ for $x=-\frac{1}{2}\ge -3$ and $x=1\ge -3$
For $x< -3\Rightarrow f''(x)=-\dfrac{3(2x^2-x-1)}{(x^2+1)^{\frac{5}{2}}}$
$f''(x)=0$ for $x=-\frac{1}{2}$ and $x=1$ but $-\frac{1}{2},1> -3$
We can see that $I_1\left(-\frac{1}{2},\frac{\sqrt{5}}{5}\right)$ and $I_2(1,2\sqrt{2})$ are inflection points.
But because $f''(x)$ is changing sign for $x<-3$ and $x\ge -3$, $I_3(-3,0)$ is also the inflection point of $f(x)$.
Is it correct that $I_3$ is one of the inflection points?
It depends on how you define inflection point. If you require the existence of the tangent, it is not; if you just require that the function changes from being convex to concave at either side of the point, it is.
Note that $f$ is not differentiable at $-3$, so the second derivative cannot exist at $-3$.