Answer
See the proof below.
Work Step by Step
$$
y=\frac{\sin x}{x}
$$
$ \Rightarrow $
$$
y^{\prime}(x)=\frac{x \cos x -\sin x}{x^{2}}
$$
$ \Rightarrow $
$$
\begin{split}
y^{\prime\prime}(x) &=\frac{-x^{2}\sin x -2x \cos x +2\sin x}{x^{3}}\\
&=\frac{(2-x^{2})\sin x -2x \cos x }{x^{3}}
\end{split}
$$
If $(x,y)$ is an infection point, then $ y^{\prime\prime}(x) =0$
$ \Rightarrow $
$$
\begin{split}
(2-x^{2})\sin x -2x \cos x &=0\\
(2-x^{2})\sin x &=2x \cos x \\
(2-x^{2})^{2}\sin^{2} x &=4x^{2} \cos^{2} x \\
(2-x^{2})^{2}\sin^{2} x &=4x^{2}(1-\sin^{2})\\
(4-4x^{2}+x^{4})\sin^{2} x &=4x^{2}-4x^{2}\sin^{2} \\
(4+x^{4})\sin^{2} x &=4x^{2} \\
(4+x^{4}) \frac{\sin^{2} x}{x^{2}} &=4 \\
(4+x^{4}) y^{2} &=4
\end{split}
$$
where $ y=\frac{\sin x}{x}$.
Thus we find that the infection points of the curve $ y=\frac{\sin x}{x}$ lie on the curve $(4+x^{4}) y^{2} =4$
