Assuming that the base of the $\log$ in the question is $e$.
Let us try to rewrite each of these functions in the form $e^{\,\small\rm something}$, to make the comparison easier.
$$\def\d{\displaystyle\,}
\begin{array}{l|ll}
a. & e^{n} / n & = \dfrac{e^{\d n}}{e^{\d(\ln n)}} & = e^{\d (n - \ln n)}\\[1em]
b. & e^{n - 0.9 \ln n} &&= e^{\d (n - 0.9 \ln n)}\\[1em]
c. & 2^n &= \left (e^{\d\ln 2} \right )^{\d n} &= e^{\d (n \ln 2)}\\[1em]
d. & (\ln n)^{n-1} &= \left (e^{\d\ln \ln n} \right )^{\d n-1} &= e^{\d (n \ln \ln n - \ln\ln n)}
\end{array}$$
Now, if we just compare the exponents of all, we can clearly see that $(n \ln \ln n - \ln \ln n)$ grows faster than the rest. Note that in option $(C)$, the multiplicative $\ln 2$ is a constant, and hence grows slower than the multiplicative $\ln \ln n$ from option $(D)$.
This implies that $e^{\d (n \ln \ln n - \ln\ln n)}$ grows the fastest, and hence, $(\ln n)^{n-1}$ grows the fastest.
Thus, option (D) is the correct answer.