Derivation of nth derivative(term) using mathmatical induction of$~y_{k}^{(k)}=(-1)^{k}x^{-(k+1)}\exp(1/x)~$

Assume $ \frac{d^k y_k}{dx^n}=(-1)^kx^{-(k+1)}\exp\left(\frac{1}{x}\right)$ holds for some $k \geq 1$.

The first step I took was split the derivative operator:

$$ \frac{d^{k+1} y_{k+1}}{dx^{k+1}} = \frac{d^k}{dx^k} \left( \frac{d}{dx} y_{k+1} \right) = \frac{d^k}{dx^k} \left( \frac{d}{dx} x^k \exp\left(\frac{1}{x}\right) \right) $$

Computing the inner derivative, we have

$$ \begin{aligned} \frac{d}{dx} x^k \exp\left(\frac{1}{x}\right) &= kx^{k-1} \exp\left(\frac{1}{x}\right) + x^k \exp\left(\frac{1}{x}\right) \left(-x^{-2}\right) \\ &= x^{k-1}\left(k-\frac{1}{x}\right) \exp\left(\frac{1}{x}\right) \end{aligned} $$

Thus,

$$ \frac{d^{k+1} y_{k+1}}{dx^{k+1}} = \frac{d^k}{dx^k} \left( x^{k-1}\left(k-\frac{1}{x}\right) \exp\left(\frac{1}{x}\right)\right) $$

We can now use the linearity of the derivative to split this up.

$$ \frac{d^{k+1} y_{k+1}}{dx^{k+1}} = \frac{d^k}{dx^k} \left( kx^{k-1}\exp\left(\frac{1}{x}\right) \right)- \frac{d^k}{dx^k} \left( x^{k-2}\exp\left(\frac{1}{x}\right) \right) $$

I believe, at this point, you may have the necessary hints to keep going. If not, please leave a comment, and I will complete my answer with the full solution. Note that you will need to use the principle of strong induction, i.e., you will need to utilize both $y_k$ and $y_{k-1}$.