Evaluate limit of the form $0^{\infty}$ $$\lim_{n\to\infty}\left(\frac{n^4-3\cdot n^3-n^2+2\cdot n-1}{n^5+n^4-n^3-3\cdot n^2-3\cdot n+1}\right)^{\frac{6\cdot n^5-2\cdot n^4-2\cdot n^3+n^2-2\cdot n}{9\cdot n^4-2\cdot n^3+n^2+3\cdot n}}$$
It is $$\lim_{n\to \infty}\left(\frac{1}{n}+o\left(\frac{1}{n^2}\right)\right)^{(n+o(1))}$$
How can I bring it to a form that I can compute?

This is one of the many cases where composition of Taylor series is useful.
We have $y_n=A_n^{B_n}$ where
$$A_n=\frac{n^4-3 n^3-n^2+2 n-1}{n^5+n^4-n^3-3 n^2-3 n+1}\quad \text{and} \quad B_n=\frac{6 n^4-2 n^3-2 n^2+n-2}{9 n^3-2 n^2+n+3}$$
We shall work with
$$\log(y_n)=B_n \log(A_n)$$
Using the long division, we have that
$$A_n=\frac{1}{n}-\frac{4}{n^2}+\frac{4}{n^3}+O\left(\frac{1}{n^4}\right)$$
$$\log(A_n)=-  \log(n)-\frac{4}{n}-\frac{4}{n^2}+O\left(\frac{1}{n^3}\right)$$
$$B_n=\frac{2 n}{3}-\frac{2}{27}-\frac{76}{243 n}-\frac{377}{2187
   n^2}+O\left(\frac{1}{n^3}\right)$$
$$\log(y_n)=-\frac{2}{3} n \log (n)+\frac{2 }{27}\log (n)-\frac{8}{3}+\frac{4 (19 \log (n)-144)}{243 n}+O\left(\frac{1}{n^2}\right)$$ Now,using the expansion to $O\left(\frac{1}{n}\right)$
$$y_n=e^{\log(y_n)}={e^{-8/3}} n^{2/27}n^{-2 n/3}=n^{\frac{2(1-9 n)}{27} }e^{-8/3}$$ as @Gary wrote in comments.
For $n=100$, the relative error is $0.97$% and for $n=1000$, it is $0.02$%.
Edit
We can make it more general for the case where
$$A_n=\frac{a}{n}+\frac{b}{n^2}+O\left(\frac{1}{n^3}\right)\quad \text{and} \quad B_n=c n+d+O\left(\frac{1}{n}\right)$$ and get
$$y_n=e^{\frac{b c}{a}} \left(\frac{a}{n}\right)^{c n+d}$$
