Show that if $y=x^{n-1}e^{1/x}$ then $D^ny=\frac{(-1)^ne^{1/x}}{x^{n+1}}$ Show that if $y=x^{n-1}e^{1/x}$ then $D^ny=\frac{(-1)^ne^{1/x}}{x^{n+1}}$
$y_1={{\rm e}^{{x}^{-1}}} \left( {x}^{n-2}n-{x}^{n-2}-{x}^{n-3} \right) $
$y_2={\frac {{{\rm e}^{{x}^{-1}}} \left(  \left( -1+ \left( n-2 \right) x
 \right)  \left( n-1 \right) {x}^{n-2}-{x}^{n-3} \left( -1+ \left( n-3
 \right) x \right)  \right) }{{x}^{2}}}
$
I want to use Leibnitz rules for successive differentiation. Please help.

What about Leibniz + induction? For $\;n=1\;$ we have
$$y_1=x^0e^{1/x}=e^{1/x}\implies y_1'=-\frac1{x^2}e^{1/x}=\frac{(-1)^1\,e^{1/x}}{x^{1+1}}\;\;\color{green}{\checkmark}$$
Assume truth up to $\;n-1\;$ ,  and now prove for $\;n\;:$$${}$$
$$\begin{align*}&y_n=x^{n-1}e^{1/x}=x\cdot x^{n-2}e^{1/x}=xy_{n-1}\implies D^n(y_n)\stackrel{\text{Leibniz}}=\sum_{k=0}^n\binom nk x^{(k)}y_{n-1}^{(n-k)}=\\{}\\
&=\binom n0xD^n(y_{n-1})+\binom n1x'D^{n-1}(y_{n-1})\stackrel{\text{Ind. Hyp.}}=xD\left(\frac{(-1)^{n-1}e^{1/x}}{x^n}\right)+n\frac{(-1)^{n-1}e^{1/x}}{x^n}=\\{}\\
&=x(-1)^{n-1}\frac{-x^{n-2}-nx^{n-1}}{x^{2n}}e^{1/x}+\frac{(-1)^{n-1}n}{x^{n}}e^{1/x}=\\{}\\
&=\require{cancel}\left(\frac{-1-nx}{x^{n+1}}+\frac n{x^n}\right)(-1)^{n-1}e^{1/x}=\frac{-1-\cancel{nx}+\cancel{nx}}{x^{n+1}}(-1)^{n-1}e^{1/x}=\frac{(-1)^ne^{1/x}}{x^{n+1}}\;\;\color{green}{\checkmark}\end{align*}$$
