In the references below, the derivative formula
\begin{equation}\label{exp-frac1x-expans}\tag{QF1}
\bigl(\operatorname{e}^{\pm1/t}\bigr)^{(n)}
=(-1)^n{\operatorname{e}^{\pm1/t}}\sum_{k=0}^{n}(\pm1)^{k}\binom{n-1}{k-1}\frac{n!}{k!}\frac1{t^{n+k}}, \quad n\ge0
\end{equation}
was derived alternatively. From \eqref{exp-frac1x-expans}, it follows that
\begin{equation}\label{exp-frac1x-alpha}\tag{QF2}
\bigl(\operatorname{e}^{\alpha/t}\bigr)^{(n)}
=(-1)^n\operatorname{e}^{\alpha/t}\sum_{k=0}^{n}\alpha^{k}\binom{n-1}{k-1}\frac{n!}{k!}\frac1{t^{n+k}}, \quad n\ge0.
\end{equation}
Taking $\alpha=-\frac{\lambda}{2}$ in \eqref{exp-frac1x-alpha} gives
\begin{equation*}
\biggl[\exp\biggl(-\frac{\lambda}{2}\frac{1}{t}\biggr)\biggr]^{(k)}
=(-1)^k\exp\biggl(-\frac{\lambda}{2}\frac{1}{t}\biggr) \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{t^{k+\ell}}, \quad k\ge0.
\end{equation*}
Therefore, in light of the Leibnitz rule for differentiation, we obtain
\begin{gather*}
\begin{aligned}
f^{(n)}(x)&=\biggl[\exp\biggl(\frac{\lambda}{\mu}\biggr) \exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr) \exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr)\biggr]^{(n)}\\
&=\exp\biggl(\frac{\lambda}{\mu}\biggr) \sum_{k=0}^{n}\binom{n}{k}\biggl[\exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr)\biggr]^{(n-k)} \biggl[\exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr)\biggr]^{(k)}
\end{aligned}\\
\begin{aligned}
&=\exp\biggl(\frac{\lambda}{\mu}\biggr) \sum_{k=0}^{n}\binom{n}{k} \exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr) \biggl(-\frac{\lambda}{2\mu^2}\biggr)^{n-k} (-1)^k\exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr) \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{x^{k+\ell}}\\
&=\exp\biggl(\frac{\lambda}{\mu}-\frac{\lambda}{2\mu^2}x-\frac{\lambda}{2}\frac{1}{x}\biggr) \sum_{k=0}^{n}\binom{n}{k} \biggl(-\frac{\lambda}{2\mu^2}\biggr)^{n-k} (-1)^k \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{x^{k+\ell}}\\
&=f(x)\frac{(-1)^nn!}{2^n\mu^{2n}x^{2n}} \sum_{k=0}^{n}\frac{\lambda^{n-k}}{(n-k)!} 2^{k}\mu^{2k} \sum_{\ell=0}^{k}\frac{(-\lambda)^{\ell}}{(2\ell)!!} \binom{k-1}{\ell-1}x^{2n-k-\ell}, \quad n\ge0.
\end{aligned}
\end{gather*}
However, numerical computation by the software Mathematica shows that the derivative formula
\begin{equation*}
\boxed{f^{(n)}(x)=f(x)\frac{(-1)^nn!}{2^n\mu^{2n}x^{2n}} \sum_{k=0}^{n}\frac{\lambda^{n-k}}{(n-k)!} 2^{k}\mu^{2k} \sum_{\ell=0}^{k}\frac{(-\lambda)^{\ell}}{(2\ell)!!} \binom{k-1}{\ell-1}x^{2n-k-\ell}, \quad n\ge0}
\end{equation*}
is not correct. But I cannot find any error in the above proof. What's wrong? How's wrong?

References
 1. S. Daboul, J. Mangaldan, M. Z. Spivey, and P. J. Taylor, *The Lah numbers and the $n$th derivative of $e^{1/x}$*, Math. Mag. **86** (2013), no. 1, 39--47; Available online at http://dx.doi.org/10.4169/math.mag.86.1.039.
 2. F. Qi, *An explicit formula for the Bell numbers in terms of the Lah and Stirling numbers*, Mediterr. J. Math. **13** (2016), no. 5, 2795--2800; available online at https://doi.org/10.1007/s00009-015-0655-7.
 3. X.-J. Zhang, F. Qi, and W.-H. Li, *Properties of three functions relating to the exponential function and the existence of partitions of unity*, Int. J. Open Probl. Comput. Sci. Math. **5** (2012), no. 3, 122--127; available online at https://doi.org/10.12816/0006128.
 4. https://qifeng618.wordpress.com/2018/05/10/some-papers-related-to-the-function-e1-x-and-the-lah-numbers/