Showing that the inflection points of $x\sin x$ lie on a certain curve 
Show that the inflection points of $f(x)=x\sin x$ are on the curve $$y^2(x^2+4)=4x^2.$$

I checked the graph of each function, but it seems that $f$ has infinitely many inflection points. How should I proceed?

$f(x)$ has an inflection point only if $f''(x)=0.$
$$f'(x) = \sin x + x \cos x$$
$$f''(x) = \cos x + \cos x -x \sin x=2\cos x - f(x)$$
$$f''(x) = 2\sqrt{1-\sin^2 (x)} - f(x)$$
$$f''(x) = 2\sqrt{1-f(x)^2/x^2} - f(x)$$
$$0=2\sqrt{1-f(x)^2/x^2} - f(x)$$
$$y=2\sqrt{1-y^2/x^2}$$
$$y^2 = 4-4y^2/x^2$$
$$...$$
