Answer
Let
$$
y=\frac{\sin x}{x}
$$
$\Rightarrow$
$$
y^{\prime}=\frac{x \cos x-\sin x}{x^{2}}
$$
$\Rightarrow$
$$
y^{\prime \prime}=\frac{-x^{2} \sin x-2 x \cos x+2 \sin x}{x^{3}}
$$
If $ (x, y)$ is an inflection point, then
$$
\begin{aligned}
y^{\prime \prime}=0 & \Rightarrow\left(2-x^{2}\right) \sin x=2 x \cos x \\
\left(2-x^{2}\right) \sin x=2 x \cos x & \Rightarrow\left(2-x^{2}\right)^{2} \sin ^{2} x=4 x^{2} \cos ^{2} x \\
& \Rightarrow \left(2-x^{2}\right)^{2} \sin ^{2} x=4 x^{2}\left(1-\sin ^{2} x\right)
\end{aligned}
$$
$ \Rightarrow$
$$
\left(4-4 x^{2}+x^{4}\right) \sin ^{2} x=4 x^{2}-4 x^{2} \sin ^{2} x
$$
$ \Rightarrow$
$$
\left(4+x^{4}\right) \sin ^{2} x=4 x^{2}
$$
$ \Rightarrow$
$$
\left(x^{4}+4\right) \frac{\sin ^{2} x}{x^{2}}=4
$$
Since $y=\frac{\sin x}{x}$ then we obtain:
$$
y^{2}\left(x^{4}+4\right)=4
$$
Work Step by Step
Let
$$
y=\frac{\sin x}{x}
$$
$\Rightarrow$
$$
y^{\prime}=\frac{x \cos x-\sin x}{x^{2}}
$$
$\Rightarrow$
$$
y^{\prime \prime}=\frac{-x^{2} \sin x-2 x \cos x+2 \sin x}{x^{3}}
$$
If $ (x, y)$ is an inflection point, then
$$
\begin{aligned}
y^{\prime \prime}=0 & \Rightarrow\left(2-x^{2}\right) \sin x=2 x \cos x \\
\left(2-x^{2}\right) \sin x=2 x \cos x & \Rightarrow\left(2-x^{2}\right)^{2} \sin ^{2} x=4 x^{2} \cos ^{2} x \\
& \Rightarrow \left(2-x^{2}\right)^{2} \sin ^{2} x=4 x^{2}\left(1-\sin ^{2} x\right)
\end{aligned}
$$
$ \Rightarrow$
$$
\left(4-4 x^{2}+x^{4}\right) \sin ^{2} x=4 x^{2}-4 x^{2} \sin ^{2} x
$$
$ \Rightarrow$
$$
\left(4+x^{4}\right) \sin ^{2} x=4 x^{2}
$$
$ \Rightarrow$
$$
\left(x^{4}+4\right) \frac{\sin ^{2} x}{x^{2}}=4
$$
Since $y=\frac{\sin x}{x}$ then we obtain:
$$
y^{2}\left(x^{4}+4\right)=4
$$
