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 $-2le xle2$, now I need to prove that there must be a point $x_{0}$ on the interval $(-2,2)$ such that $f^{prime prime}left(x_{0}right)+3left(fleft(x_{0}right)right)=0$.

```
(Series(1/2 (f'(x))^2 + f(x)^3, {x, 0, 1})) // FullSimplify
1/2 (Series(f'(x), {x, 0, 1}) // Normal)^2 + (Series(
f(x), {x, 0, 1}) // Normal)^3 // Expand
(Series(f''(x), {x, 0, 1}) // Normal)^2 + (3*Series(f(x), {x, 0, 1}) //
Normal)^2 // Expand
```

The above code does not reveal the nature of the problem and solve it cleverly. What can I do to solve this problem?