Since, $\alpha$ and $\beta$ are the roots of the equation $x^{2}-2 x-1=0$, then
Sum of roots, $\alpha+\beta=2$ and
product of the roots $\alpha \beta=-1$
Since, $(\alpha+\beta)=\alpha^{2}+\beta^{2}+2 \alpha \beta$
$\Rightarrow 4=\alpha^{2}+\beta^{2}-2$
$\Rightarrow \alpha^{2}+\beta^{2}=6$
Now, $\alpha^{2} \beta^{-2}+\alpha^{-2} \beta^{2}=\frac{\alpha^{2}}{\beta^{2}}+\frac{\beta^{2}}{\alpha^{2}}=\frac{\alpha^{4}+\beta^{4}}{(\alpha \beta)^{2}}$
$\Rightarrow \left(\alpha^{2}+\beta^{2}\right)^{2}=6^{2}$
$\Rightarrow \alpha^{4}+\beta^{4}+2 \alpha^{2} \beta^{2}=36$
$\Rightarrow \alpha^{4}+\beta^{4}+2=36$
$\Rightarrow \alpha^{4}+\beta^{4}=34 \ldots$ (i)
$\Rightarrow \frac{\alpha^{4}+\beta^{4}}{(\alpha \beta)^{2}}=\frac{34}{(-1)^{2}}=34$
[Putting value of $\alpha^{4}+\beta^{4}=34$ from Equation (i)].