Want help with proving a calculus theory Let $f(x)$ have a second derivative on the closed interval $[-2,2]$. If   $\left| f(x) \right| \le 1$ and $\frac{1}{2}  (f^{\prime}(0))^2+f(0)^3>\frac{3}{2}  $ when $-2\le x\le2$, now I need to prove that there must be a point $x_{0}$ on the interval $(-2,2)$ such that $f^{\prime \prime}\left(x_{0}\right)+3f\left(x_{0}\right)^2=0$.
(Series[1/2 (f'[x])^2 + f[x]^3, {x, 0, 1}]) // FullSimplify

The above method does not reveal the nature of the problem and solve it cleverly. I want to use a more generic and heuristic method to verify the conclusion of this abstract function problem. What can I do to solve this problem?
The source of this problem (张宇高等数学18讲):

 A: First note that $\ \displaystyle\frac{f'(0)^2}{2}+f(0)^3>\frac{3}{2}\ $ and $\ |f(x)|\le1\ $ for $\ x\in[-2,2]\ $ implies that
$$
|f'(0)|>1\ .
$$
By the mean-value theorem, there must exist $\ \xi_1\in(-2,0)\ $ and $\ \xi_2\in(0,2)\ $ such that $\ \frac{f(0)-f(-2)}{2}=f'\left(\xi_1\right)\ $ and $\ \frac{f(2)-f(0)}{2}$$=f'(\xi_2)\ $. It follows that
\begin{align}
\left|f'\left(\xi_1\right)\right|&\le\frac{|f(-2)|+|f(0)|}{2}\\
&\le1\ ,\\
\left|f'\left(\xi_2\right)\right|&\le\frac{|f(0)|+|f(2)|}{2}\\
&\le1\ \text{, and}\\
\frac{f'(\xi_i)^2}{2}+ f(\xi_i)^3&\le\frac{1}{2}+1\\
&\le\frac{3}{2}\ \text{ for }\ i=1,2\ .
\end{align}
Thus, since $\ \displaystyle\frac{f'(x)^2}{2}+f(x)^3\ $ is differentiable on $\ (\xi_1,\xi_2)\ $, continuous on $\ [\xi_1,\xi_2]\ $, $\ \displaystyle\frac{3}{2}\ \ge \frac{f'(\xi_1)^2}{2}+f(\xi_1)^3\ $, $\ \displaystyle\frac{3}{2}\ \ge \frac{f'(\xi_2)^2}{2}+f(\xi_2)^3\ $, and $\ \displaystyle\frac{f'(0)^2}{2}+f(0)^3>\frac{3}{2}\ $, it must achieve its supremum over the interval $\ [\xi_1,\xi_2] \ $ at some point $\ x_0\in(\xi_1,\xi_2)\ $ in the interval's interior, where its derivative must vanish:
\begin{align}
0&=f'\left(x_0\right)f''(x_0)+3f(x_0)^2f'(x_0)\\
&=f'(x_0)\left(f''(x_0)+3f(x_0)^2\right)\ .
\end{align}
But since $\ \frac{f'\left(x_0\right)^2}{2}>$$\frac{3}{2}-f\left(x_0\right)^3\ge$$\frac{1}{2}\ $, it follows that
$$
f''(x_0)+3f(x_0)^2=0\ .
$$
Acknowldgement: This proof incorporates a simplification suggested by Martin R in the comments.
