= Solution
The first three <derivatives> factor as
$$
\boxed{
f'(x)=2x^3(3x^2-10),\quad
f''(x)=30x^2(x^2-2),\quad
f'''(x)=120x(x^2-1).
}
$$
Let $a=\sqrt{10/3}$. Sign analysis of the factors gives
$$
\begin{aligned}
f'(x)>0&\iff x\in(-a,0)\cup(a,\infty),\\
f''(x)>0&\iff x\in(-\infty,-\sqrt2)\cup(\sqrt2,\infty),\\
f'''(x)>0&\iff x\in(-1,0)\cup(1,\infty).
\end{aligned}
$$
All displayed endpoints are excluded because a factor vanishes there; the extra zero of $f''$ at $0$ does not create a positive <interval>.
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