<s>[INST] $n$-th derivative of $y$ in $x^y=e^{x-y}$ Question:

If $x^y=e^{x-y}$, find a general formula for $\displaystyle\frac{d^ny}{dx^n}$.

My Attempt:
$$x^y=e^{x-y}$$
$$\Rightarrow\ \ \ \ \ y\ln x=x-y$$
$$\Rightarrow\ \ \ \ \ y=\frac{x}{\ln x+1}$$
So
$$\begin{align}\frac{dy}{dx}&=\frac{\ln x+1-1}{(\ln x+1)^2}\\
&=\frac{\ln x}{(\ln x+1)^2}
\end{align}$$
Then
$$\begin{align}\frac{d^2y}{dx^2}&=\frac{\frac{1}{x}(\ln x+1)^2+\frac{2\ln x}{x}(\ln x+1)}{(\ln x+1)^4}
\end{align}$$
Now, I can see that the denominator satisfies a straight-forward pattern $(\ln x+1)^{2^n}$. So the general formula for $\displaystyle\frac{d^ny}{dx^n}$ is
$$\frac{d^ny}{dx^n}=\frac{g(n)}{(\ln x+1)^{2^n}}$$But I can't derive any formula for the numerator $g(n)$, nor can I see any regular pattern in the structure. Should I differentiate the initial expression $x^y=e^{x-y}$ implicity? Can anyone help me in that?
 [/INST] Here is a formula of the $n$-th derivative of $\frac{f}{g}$  in terms of  derivatives  of $f$  and $g$ which might be helpful.

Let $D_x$ represent differentiation with respect to $x$. Hence $D^n_x f(x)$ is the $n$-th derivative of $f$ with respect to $x$.  The following holds true for $n$ times differentiable functions $f$ and $g$
  \begin{align*}
D_x^n\left(\frac{f}{g}\right)=\sum_{k=0}^n\sum_{j=0}^{k} (-1)^j\binom{n}{k}\binom{k+1}{j+1}\frac{1}{g^{j+1}}
D_x^{n-k}\left(f\right) D_x^{k}\left( g^j\right)\tag{1}
\end{align*}
This formula is based upon the Hoppe Form of Generalized Chain Rule and a proof for it is given in this MSE post.

In the current situation we have $f(x)=x$ and $g(x)=\ln(x)+1$ and we obtain from (1) if we change the order of summation of the outer sum by replacing $k$ with $n-k$:

\begin{align*}
\color{blue}{D_x^n\left(\frac{x}{\ln x+1}\right)}
&=\sum_{k=0}^n\sum_{j=0}^{n-k}(-1)^j\binom{n}{k}\binom{n-k+1}{j+1}
\frac{1}{(\ln x + 1)^{j+1}}D_x^k(x)D_x^{n-k}\left((\ln x + 1)^j\right)\\
&=\color{blue}{\sum_{j=0}^{n}(-1)^j\binom{n+1}{j+1}
\frac{x}{(\ln x+1)^{j+1}}D_x^{n}\left((\ln x+1)^j\right)}\\
&\qquad\color{blue}{+\sum_{j=0}^{n-1}(-1)^jn\binom{n+1}{j+1}
\frac{1}{(\ln x+1)^{j+1}}D_x^{n-1}\left((\ln x+1)^j\right)}\\
\end{align*}
  Since $D_x^k(x)=0$ if $k>1$ it is sufficient to consider the first two terms $k=0,1$ of the outer sum.

Let's look at a small example in order to see the  formula in action
Example: $D^1_x\left(\frac{x}{\ln x + 1}\right)$
\begin{align*}
\color{blue}{D_x^1\left(\frac{x}{\ln x +1}\right)}
&=\sum_{j=0}^1(-1)^j\binom{2}{j+1}\frac{x}{(\ln x+1)^{j+1}}D_x^1\left((\ln x+1)^j\right)\\
&\qquad +\sum_{j=0}^0(-1)^j\binom{1}{j+1}\frac{1}{(\ln x+1)^{j+1}}D_x^0\left((\ln x+1)^j\right)\\
&=(-1)^0\binom{2}{1}\frac{x}{\ln x+1}D_x(1)+(-1)^1\binom{2}{2}\frac{x}{(\ln x+1)^2}D_x(\ln x + 1)\\
&\qquad +(-1)^0\binom{1}{1}\frac{1}{\ln x+1}D_x^0(1)\\
&=0-\frac{x}{(\ln x+1)^2}\cdot \frac{1}{x}+\frac{1}{\ln x+1}\\
&\color{blue}{=\frac{\ln x}{(\ln x+1)^2}}
\end{align*}
in accordance with OPs result.
</s>