Prove by induction. Assume $n$ is a positive integer, $x \neq 0$ and that all derivatives exists.
$\frac{d^{n}}{dx^{n}}\Big[x^{n-1}.f\big(\frac{1}{x}) \Big]=\frac{(-1)^{n}}{x^{n+1}}.f^{n}\big(\frac{1}{x}\big)$
I did the base case $n=1$
L.H.S= $\frac{d}{dx}\Big[x^{0}.f\big(\frac{1}{x}\big)\Big]=\frac{-1}{x^{2}}f'(\frac{1}{x})$
R.H.S=$\frac{(-1)^{1}}{x^{2}}.f'(\frac{1}{x})$
Thus, the R.H.S=L.H.S. We have proved it is true for $n=1$
Now, I want to show that
$\frac{d^{n+1}}{dx^{n+1}}\Big[x^{n}.f\big(\frac{1}{x}) \Big]=\frac{(-1)^{n+1}}{x^{n+2}}.f^{(n+1)}\big(\frac{1}{x}\big)$
So,
L.H.S=$\frac{d^{n+1}}{dx^{n+1}}\Big[x^{n}.f\big(\frac{1}{x}) \Big]$ = $\frac{d^{n}}{dx^{n}}(\frac{d}{dx}x^{n}f(\frac{1}{x}))=\frac{d^{n}}{dx^{n}}(-x^{n-2}f'(\frac{1}{x})+nx^{n-1}f(\frac{1}{x}))=n\frac{d^{n}}{dx^{n}}(x^{n-1}.f(\frac{1}{x}))-\frac{d^{n}}{dx^{n}}(x^{n-2}f'(\frac{1}{x}))$
Then, since $\frac{d^{n}}{dx^{n}}\Big[x^{n-1}.f\big(\frac{1}{x}) \Big]=\frac{(-1)^{n}}{x^{n+1}}.f^{n}\big(\frac{1}{x}\big)$
So, we would have the LHS=$ n\frac{(-1)^{n}}{x^{n+1}}.f^{n}\big(\frac{1}{x}\big)-\frac{d^{n}}{dx^{n}}(x^{n-2}f'(\frac{1}{x}))$
From here I did not know what to do, would anyone have an idea how would I continue. Thank you !
The induction step works fine if you split $\frac{d^{n+1}}{dx^{n+1}}$ as $\frac{d}{dx}\frac{d^n}{dx^n}$, but you need the Leibniz rule for the starred step below:
$$\begin{align*} \frac{d^{n+1}}{dx^{n+1}}\left(x^nf\left(\frac1x\right)\right)&=\frac{d}{dx}\left(\frac{d^n}{dx^n}\left(x^nf\left(\frac1x\right)\right)\right)\\ &=\frac{d}{dx}\left(\frac{d^n}{dx^n}\left(x\cdot x^{n-1}f\left(\frac1x\right)\right)\right)\\ &\overset{*}=\frac{d}{dx}\left(\sum_{k=0}^n\binom{n}k\frac{d^k}{dx^k}(x)\frac{d^{n-k}}{dx^{n-k}}\left(x^{n-1}f\left(\frac1x\right)\right)\right)\\ &=\frac{d}{dx}\left(x\frac{d^n}{dx^n}\left(x^{n-1}f\left(\frac1x\right)\right)+n\frac{d^{n-1}}{dx^{n-1}}\left(x^{n-1}f\left(\frac1x\right)\right)\right)\\ &=\frac{d}{dx}\left(\frac{(-1)^n}{x^n}f^{(n)}\left(\frac1x\right)+n\frac{d^{n-1}}{dx^{n-1}}\left(x^{n-1}f\left(\frac1x\right)\right)\right)\\ &=\frac{(-1)^{n+1}}{x^{n+2}}f^{(n+1)}\left(\frac1x\right)+\frac{(-1)^{n+1}n}{x^{n+1}}f^{(n)}\left(\frac1x\right)+n\frac{d^n}{dx^n}\left(x^{n-1}f\left(\frac1x\right)\right)\\ &=\frac{(-1)^{n+1}}{x^{n+2}}f^{(n+1)}\left(\frac1x\right)+\frac{(-1)^{n+1}n}{x^{n+1}}f^{(n)}\left(\frac1x\right)+\frac{(-1)^nn}{x^{n+1}}f^{(n)}\left(\frac1x\right)\\ &=\frac{(-1)^{n+1}}{x^{n+2}}f^{(n+1)}\left(\frac1x\right) \end{align*}$$
All but the first two terms of the summation in the third line disappear.