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?
 A: 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.
A: $f(x)$ has an inflection point only if $f''(x)=0.$
$$f'(x) = \sin x + x \cos x$$
$$f''(x) = \cos x + \cos x -x \sin x=2\cos x - f(x)$$
$$f''(x) = 2\sqrt{1-\sin^2 (x)} - f(x)$$
$$f''(x) = 2\sqrt{1-f(x)^2/x^2} - f(x)$$
$$0=2\sqrt{1-f(x)^2/x^2} - f(x)$$
$$y=2\sqrt{1-y^2/x^2}$$
$$y^2 = 4-4y^2/x^2$$
$$...$$
A: $f(x)=x\sin x \implies f''(x)=2\cos x-x\sin x$ let $x=h$ be the point of inflexion then $f''(h)=0 \implies h=2\cot h ~~~(1)$, The point $(h,k)$ falls on the curve hence $k=f(h)= h \sin h~~~(2)$ from (1) and (2) the point $(h,k)$ will satisfy ans algebraic equation when we eliminiate the periodic function from (1) and (2).
From (2) we have $\cos h=\frac{h}{\sqrt{4+h^2}}~~~(3)$ squaring ans adding (1) and (2) we get $$\frac{h^2}{4+h^2}+\frac{k^2}{h^2}=1 \implies k^2(h^2+4)=4h^2$$
So the required curve ids $y^2(x^2+4)=4x^2$
A: Inflection points indeed check OK, as three concurrent plots should, when plotted together.
$$y - x \sin x =0\tag1$$
$$ y^{''}=2 \cos x - x \sin x=0 \tag2 $$
$$ \text{ From(2) solve for x from the equation} (\frac{2}{x}=\tan x ) \tag3 $$
The above is a transcendental equation which has roots that can be solved for by  Newton-Raphson iteration etc. The roots of green curve are directly above or below point of inflection in graphical visual verification below.
Eliminate $x \sin x $ between (1) and (2)
$$ y = 2 \cos x \tag 4 $$
This is a trigonometric equation, not transcendental. ( Not asked for by OP).
Eliminate $ \sin x $ between (1) and (2)
$$ y =\pm  \frac{2 x }{ \sqrt{4+x^2 }}\tag 5 $$
This is an algebraic equation, not transcendental.
In (4) and (5) although we have not eliminated $y$ totally , we however find that they provide valuable plot sources for inflection point loci.
The same is graphically shown below as three plots concur.
In passing btw it should be mentioned that the procedure is so nice. It even  enables finding an envelope through maxima/minima while eliminating between the curve and its first derivative.

A: The inflection points occur at
$$f''(x)=2\cos x-x\sin x=0$$ or $$y=2\cos x^*.$$
So we indeed have
$$x^2y^2+4y^2=4x^2\cos^2x+4x^2\sin^2x=4x^2.$$
$^*$This is another locus of the inflection points (magenta).

