#### Answer

$$ x=\pm 2/\sqrt{3}$$

#### Work Step by Step

Given $$y=\frac{x^{2}}{x^{2}+4}$$ Since \begin{align*} y'&=\frac{\frac{d}{dx}\left(x^2\right)\left(x^2+4\right)-\frac{d}{dx}\left(x^2+4\right)x^2}{\left(x^2+4\right)^2}= \frac{8x}{\left(x^2+4\right)^2} \\ y''&=8\cdot \frac{\frac{d}{dx}\left(x\right)\left(x^2+4\right)^2-\frac{d}{dx}\left(\left(x^2+4\right)^2\right)x}{\left(\left(x^2+4\right)^2\right)^2}=\frac{8\left(-3x^2+4\right)}{\left(x^2+4\right)^3} \end{align*}
Then $f''(x)=0$ for $x=\pm 2/\sqrt{3}$ and
$$ y''\gt 0\ \ \text{ for } -\frac{2}{\sqrt{3}}\lt x\lt \frac{2}{\sqrt{3}}\\
y''\lt 0 \ \ \text{ for } \ |x| \geq \frac{2}{\sqrt{3}}
$$
Then $f(x) $ has infection points at $ x=\pm 2/\sqrt{3}$