Given that
\[\cos 2 \theta = \frac{1 + \sqrt{5}}{4},\]find $\tan^2 \theta \tan^2 3 \theta.$
We have that
\[\cos^2 \theta - \sin^2 \theta = \frac{1 + \sqrt{5}}{4}.\]Then
\[\frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta + \sin^2 \theta} = \frac{1 + \sqrt{5}}{4},\]so
\[\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 + \sqrt{5}}{4}.\]Isolating $\tan^2 \theta,$ we find
\[\tan^2 \theta = \frac{\sqrt{5} - 2}{\sqrt{5}}.\]Then
\begin{align*}
\tan^2 3 \theta &= (\tan 3 \theta)^2 \\
&= \left( \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right)^2 \\
&= \tan^2 \theta \cdot \left( \frac{3 - \tan^2 \theta}{1 - 3 \tan^2 \theta} \right)^2 \\
&= \frac{\sqrt{5} - 2}{\sqrt{5}} \cdot \left( \frac{3 - \frac{\sqrt{5} - 2}{\sqrt{5}}}{1 - 3 \cdot \frac{\sqrt{5} - 2}{\sqrt{5}}} \right)^2 \\
&= \frac{\sqrt{5} - 2}{\sqrt{5}} \cdot \left( \frac{2 \sqrt{5} + 2}{-2 \sqrt{5} + 6} \right)^2 \\
&= \frac{\sqrt{5} - 2}{\sqrt{5}} \cdot \left( \frac{\sqrt{5} + 1}{-\sqrt{5} + 3} \right)^2 \\
&= \frac{\sqrt{5} - 2}{\sqrt{5}} \cdot \left( \frac{(\sqrt{5} + 1)(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})} \right)^2 \\
&= \frac{\sqrt{5} - 2}{\sqrt{5}} \cdot \left( \frac{8 + 4 \sqrt{5}}{4} \right)^2 \\
&= \frac{\sqrt{5} - 2}{\sqrt{5}} \cdot (2 + \sqrt{5})^2,
\end{align*}so
\begin{align*}
\tan^2 \theta \tan^2 3 \theta &= \left( \frac{\sqrt{5} - 2}{\sqrt{5}} \right)^2 (2 + \sqrt{5})^2 \\
&= \left( \frac{(2 + \sqrt{5})(2 - \sqrt{5})}{\sqrt{5}} \right)^2 \\
&= \boxed{\frac{1}{5}}.
\end{align*}