Proving an inflection point of a certain function 
Show that the inflection points of the curve $y = \sin(x)/x$ lie on the curve $y^2(x^4+4)=4$

So, I know that to find the inflection points I have to set the second derivative of $f$ = $0$
I got $f''(x) = -((x^2-2)\sin(x)+2x\cos(x))/x^3 = 0$ 
Thus, $f''(x) = 2x\cos(x) = (x^2-2)\sin(x)$ 
What got me confused is that no matter how many times I tried, I could not get $y^2(x^4+4) = 4$ from that equation.

You know $2x\cos x = (x^2-2)\sin x$ and $y=\sin(x)/x$. Taking the square of the first implies $4x^2\cos(x)^2 = (x^2-2)^2\sin(x)^2$ and thus
$$ 4x^2(1-\sin(x)^2) = (x^2-2)^2\sin(x)^2$$
and thus
$$ \sin(x)^2 = \frac{4x^2}{(x^2-2)^2+4x^2} = \frac{4x^2}{x^4+4}$$
So $y^2(x^4+4) = \frac{\sin(x)^2}{x^2} (x^4+4) = \frac{4x^2}{x^2(x^4+4)}(x^4+4)=4$.
