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?

An inflection point is where $f''$ changes sign,
so we need to find where $f''(x)=0$.
$f(x)=x\sin(x),
f'(x)=\sin(x)+x\cos(x),
f''(x)=\cos(x)+\cos(x)-x\sin(x)
=2\cos(x)-x\sin(x)
$.
So we want to show
that this satisfies
$y^2(x^2+4)=4x^2$.
If $f''(x)=0$ then
$2\cos(x)=x\sin(x)
$
or $x=2/\tan(x)$.
Since $f(x)=x\sin(x)$,
we want
$x^2\sin^2(x)(4/\tan^2(x)+4)=4x^2
$
or
$\sin^2(x)(1/\tan^2(x)+1)=1
$
or
$1=\sin^2(x)(\cos^2(x)/\sin^2(x)+1)
$
or
$1=\cos^2(x)+\sin^2(x)
$
which is true.
