I'm not sure if this method is valid so I'd like some feedback.
$$L=\lim_{n\to\infty}\frac{3^{2^n}}{2^{3^n}}=\frac{\infty}{\infty}=\lim_{n\to\infty}\frac{\frac{\mathrm d}{\mathrm dn}(3^{2^n})}{\frac{\mathrm d}{\mathrm dn}(2^{3^n})}=\lim_{n\to\infty}\frac{2^n3^{2^n}}{3^n2^{3^n}}=\lim_{n\to\infty}(\frac{2}{3})^n\cdot\lim_{n\to\infty}\frac{3^{2^n}}{2^{3^n}}=L\cdot\lim_{n\to\infty}(\frac{2}{3})^n$$ $\lim_{n\to\infty}(\frac{2}{3})^n=0$ so $L=L\cdot0\implies L=0$.
Let's not presuppose the limit exists. Instead note $$\ln\frac{3^{2^n}}{2^{3^n}}=2^n\ln 3-3^n\ln 2=O(2^n)-O(3^n)=-O(3^n)\\\implies\lim_{n\to\infty}\ln\frac{3^{2^n}}{2^{3^n}}=-\infty\implies\lim_{n\to\infty}\frac{3^{2^n}}{2^{3^n}}=\exp-\infty=0.$$