The possible inflection points are where \(f''(x)\) is undefined or equal to 0.
Based on the factors of \(f''(x)\), possible points of inflection are \(0\), \(-3\), and \(3\).
We can test intervals:
$$ \begin{array}{|c|c|c|c|c|} \hline
\text{Interval} & -\infty \lt x \lt -3 & -3 \lt x \lt 0 & 0 \lt x \lt 3 & 3\lt x \lt \infty \\ \hline
\text{ Test Value} & x=-4 & x=-2 & x= 2 & x= 4\\ \hline
\text{ Sign of }f''(x) & f''(-4) \gt 0 & f''(-2) \gt 0 & f''(2) \lt 0 & f''(4) \gt 0 \\ \hline
\end{array} $$
The sign changes at \(x=-3\) and \(x=0\) only.
The graph of a function only crosses the solutions when the multiplicity is odd. If even, the graph bounces off of the \(x\)-intercept.
$$f''(x)=\colorbox{aqua}{$x^1$}(x+3)^2\colorbox{aqua}{$(x-3)^3$} $$
If the graph passes through the \(x\)-axis, it must change sign. For the graph of \(f''(x)\), this corresponds to an inflection point.