How can I solve $\lim\limits_{n \to \infty} \frac{n \log_2 \log_2 n}{3^{\log_2 n^2}}= 0$ How can I solve the flowing limit?

$$\lim\limits_{n \to \infty} \frac{n \log_2( \log_2 n)}{3^{\log_2 n^2}}=0$$

Attempt 1:
$\log_2 n = m \implies \log_2 n^2 = 2\cdot\log_2 n = 2\cdot m$
$$\lim\limits_{n \to \infty} \frac{n \log_2( \log_2 n)}{3^{\log_2 n^2}}=\lim\limits_{n \to \infty} \frac{n \log_2(m)}{3^{2 \cdot m}}$$
$\log_2(m) = r \iff 2^r = m$
$$\lim\limits_{n \to \infty} \frac{n \log_2(m)}{3^{2 \cdot m}} = \lim\limits_{n \to \infty} \frac{n\cdot 2^r}{3^{2 \cdot m}}$$
$m = \log_2n \iff2^m=n$
$$\lim\limits_{n \to \infty} \frac{n\cdot 2^r}{3^{2 \cdot m}} = \lim\limits_{n \to \infty} \frac{2^m\cdot 2^r}{3^{2 \cdot m}} = \lim\limits_{n \to \infty} \frac{2^{r+m}}{9^{m}} =0$$

I'm not sure if it is so right. 
Are there other ways of solve this limit?
Thanks!

Attempt 2 (By  L'Hopital's rule):
Let $f(n) = 3^{\log_2 n^2}$ and $g(n) = n \log_2( \log_2 n) \implies \lim\limits_{n \to \infty} f(n) = \infty \land \lim\limits_{n \to \infty} g(n) = \infty$
$$\lim\limits_{n \to \infty} \frac{g(n)}{f(n)}\implies \lim\limits_{n \to \infty} \frac{g´(n)}{f´(n)}$$
I get:
$f´(n)= \frac{dn}{n}(3^{\log_2 n^2}) = \frac{ 3^{\log_2 n^2} 2 \log(3)}{n \log(2)}$
$g´(n)= \frac{dn}{n}(n \log_2( \log_2 n)) = \frac{\log(n)\log(\log_2n)+1}{\log(2) \log(n)}$
also
$$\lim\limits_{n \to \infty} \frac{g´(n)}{f´(n)} = \lim\limits_{n \to \infty} \frac{\frac{\log(n)\log(\log_2n)+1}{\log(2) \log(n)}}{\frac{ 3^{\log_2 n^2} 2 \log(3)}{n \log(2)}} = \lim\limits_{n \to \infty} \frac{(\log(n)\log(\log_2n)+1)\cdot(n \log(2))}{(\log(2) \log(n))\cdot(3^{\log_2 n^2} 2 \log(3))}$$
This does not help me :( 
Which is the most optimal form to solve this limit?
 A: $\lim\limits_{n \to \infty} \frac{n \log_2( \log_2 n)}{3^{\log_2 n^2}}=0
$
We have
$\begin{array}\\
3^{\log_2 n^2}
&=e^{\ln 3\log_2 n^2}\\
&=e^{2\ln 3\log_2 n}\\
&=e^{2\ln 3 \ln(n)/\ln 2}\\
&=e^{c \ln(n)}
\qquad\text{where }c = 2\ln 3/\ln 2 > 1\\
&=(e^{ \ln(n)})^c\\
&= n^c\\
\end{array}
$
so
$\frac{n \log_2( \log_2 n)}{n^c}
=\frac{\log_2( \log_2 n)}{n^{c-1}}
\to 0
$
since
$\frac{\ln(n)}{n^a}
\to 0$
for any $a > 0$.
A: Let $u=\log_{2}(n)$ and notice that $n=2^u$. $$\lim_{n\to\infty}\frac{n\log_{2}(\log_{2}(n))}{3^{\log_{2}(n^2)}}=\lim_{n\to\infty}\frac{n\log_{2}(\log_{2}(n))}{3^{2\log_{2}(n)}}=$$ $$\lim_{u\to\infty}\frac{2^u\log_{2}(u)}{3^{2u}}=\lim_{u\to\infty}\frac{2^u\log_{2}(u)}{9^u}=$$ $$\lim_{u\to\infty}(\frac{2}{9})^u\log_{2}(u)=\lim_{u\to\infty}\frac{\ln(u)}{\ln(2)(\frac{9}{2})^u}$$
Now we get an indeterminate form of $\frac{\infty}{\infty}$, so we can apply L'Hospital's Rule.
$$\lim_{u\to\infty}\frac{\ln(u)}{\ln(2)(\frac{9}{2})^u}=\lim_{u\to\infty}\frac{\frac{d}{du}(\ln(u))}{\ln(2)(\frac{d}{du}(\frac{9}{2})^u)}=$$ $$\lim_{u\to\infty}\frac{\frac{1}{u}}{\ln(2)\ln(\frac{9}{2})(\frac{9}{2})^u}=\lim_{u\to\infty}\frac{1}{\ln(2)\ln(\frac{9}{2})(\frac{9}{2})^uu}=0$$
So, $$\lim_{n\to\infty}\frac{n\log_{2}(\log_{2}(n))}{3^{\log_{2}(n^2)}}=0$$
