If $a_n$ is a sequence, then $\lim_{n\to\infty}a_n = 0$ if and only if $\lim_{n\to\infty} |a_n| = 0$. So if the absolute value of the terms does not go to zero, then the terms don't go to zero.
In other words, if you know that $\lim \frac{n}{\ln (n)}\neq 0$, then it follows that $\lim\frac{(-1)^n n}{\ln n}\neq 0$. (because the former is the limit of the absolute values of the latter). So if your reduction is accurate, and your proof that the limit without the $(-1)^n$ factor is not zero is accurate, then you are done.
That said, I'm not sure how you manage to get to that limit. If you had $(\ln(n^n))^{-1}$, then I would agree; but how do you manage to get rid of the exponent and cancel an $n$ if you have
\[ \frac{(-1)^n n^2}{(\ln n)^n} \text{?}\]
Added: Well, looks like the beans have been spilt; your "few algebraic manipulations" were incorrect. Remember that while $\ln(a^k) = k\ln(a)$, it is not true that $(\ln a)^k = k\ln a$. $\ln(a^k)$ is the logarithm of the $k$th power of $a$; $(\ln a)^k$ is the $k$th power of the logarithm of $a$; they are usually different.