Successive differentiation of an implicit function If $y^{1/m}+y^{-1/m}=2x$, prove that $$(x^2-1)y'''+3xy''+(1-m^2)y'=0$$
I used brute force(kept on applying Product and Quotient rule) and after almost 3 pages of nasty calculations was able to get to the result. I am looking for a faster and better approach. Thanks.

Let $$y^{\frac 1m}=e^u\Rightarrow u=\frac 1m \ln y\Rightarrow u'=\frac 1m\left(\frac{y'}{y}\right)$$
Then we also have $$\cosh u =x\Rightarrow u = \operatorname{arcosh} x\Rightarrow u'=(x^2-1)^{-\frac 12}$$
We can rearrange this and square both sides so that $$(x^2-1)(u')^2=1\Rightarrow (x^2-1)\left(\frac 1m\left(\frac{y'}{y}\right)\right)^2=1$$
Now rearrange this as $$(x^2-1)(y')^2=m^2y^2$$ and differentiate:
$$2x(y')^2+(x^2-1)2y'y''=2m^2yy'$$
Finally, cancel $2y'$ and differentiate again:
$$y'+xy''+2xy''+(x^2-1)y'''=m^2y'$$
And thus we are done.
