Assuming you want a confidence interval for $\theta$, you may use the CLT or a known chi distribution for variances and some integrals to get an interval for the parameter. First, we have that
$$
\hat{\theta}=-\frac{n}{\sum ln\ x_i}.
$$
For the expected value we have the integral
$$
E[X] = \int_0^1 \theta x^{(\theta-1)}x\ \text{d}\theta = \frac{\theta}{1+\theta}
$$
And for the second moment
$$
\mu_2 = \int_0^1 \theta x^{\theta-1}x^2\ \text{d}\theta = \frac{\theta}{2+\theta}
$$
So the variance is
$$
Var(X) = \frac{\theta}{2+\theta} - \left( \frac{\theta}{1+\theta} \right)^2 = \\
= \frac{\theta}{(\theta + 1)^2(\theta+2)}.
$$
Now then, we know that the following quotient has a chi-squared distribution, so in this case
$$
C = \frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1}
\text{ and } C = \frac{(n-1)S^2}{\frac{\theta}{(\theta + 1)^2(\theta+2)}}
$$
where $S^2 = \frac{1}{n-1}\sum(X_i-\overline{X})^2$. Finally we can work with the random variable $C$ distributed as a chi squared. We want a confidence interval of $(1-\alpha)$. Then
$$
P \left( \chi^2_{n-1}(\alpha/2) < C < \chi^2_{n-1}(1-\alpha/2)\right) = 1-\alpha
$$
denoting that $\chi^2_{n-1}(\beta)$ is the $\beta$ percentile of the $\chi^2$ which may be found in reference tables.
From here we work with the interval only substituting the percentiles with $\frac{1}{D_l}$ and $\frac{1}{D_s}$, lower and superior.
$$
\frac{1}{D_l} < C < \frac{1}{D_s} \implies
\frac{1}{D_l} < \frac{(n-1)S^2}{\frac{\theta}{(\theta + 1)^2(\theta+2)}} < \frac{1}{D_s} \\\ \\\ \\\
D_l > \frac{\frac{\theta}{(\theta + 1)^2(\theta+2)}}{(n-1)S^2} > D_s \\\ \\\ \\\
D_l((n-1)S^2) > \frac{\theta}{(\theta + 1)^2(\theta+2)} > D_s((n-1)S^2) \\\ \\\ \\\
(D_l((n-1)S^2)) > \frac{\theta}{(\theta + 1)^2(\theta+2)} > (D_s((n-1)S^2)) \\\ \\\ \\\
(\hat{\theta} + 1)^2(\hat{\theta}+2)\left( D_l((n-1)S^2) \right)
> \theta >
(\hat{\theta}+ 1)^2(\hat{\theta}+2)\left( D_s((n-1)S^2) \right) \\\ \\\ \\\
$$
where, taking the before definition and derivation, $\hat{\theta} = -\frac{n}{\sum ln\ x_i}$ and $S^2 = \frac{1}{n-1}\sum(X_i-\overline{X})^2$. When we sent the $\theta$ to the other side, we simply used its estimate to work around the problem. (However a mathematician should confirm this is valid!) Either way, if you find a better known distribution that may include $\theta$, then try it and tell us if it is simpler!
Hope it helps!
Note: made some edits due to an error :/
First note that $Z_{0.025}=-Z_{0.975}=-1.96$ and solve the inequaalities
$$-1.96 \le \frac{(\overline{Y} - 1.5\theta)\sqrt{12n}}{5\theta} \le 1.96$$
Multiplying by $5\theta$ and dividing by $\sqrt{12n}$, obtain
$$-\frac{1.96\cdot 5\theta}{\sqrt{12n}} \le \overline{Y} - 1.5\theta \le \frac{1.96\cdot 5\theta}{\sqrt{12n}}$$
Add $1.5\theta$ to all parts:
$$-\frac{1.96\cdot 5\theta}{\sqrt{12n}}+1.5\theta \le \overline{Y} \le \frac{1.96\cdot 5\theta}{\sqrt{12n}}+1.5\theta$$
Take $\theta$ out of brackets:
$$\theta\left(\frac32-\frac{9.8}{\sqrt{12n}}\right) \le \overline{Y} \le \theta\left(\frac32+\frac{9.8}{\sqrt{12n}}\right)$$
Solving both inequalities with respect to $\theta$ for sufficiently large $n$ which guarantee both brackets are positive, we have:
$$\frac{\overline{Y}}{\frac32+\frac{9.8}{\sqrt{12n}}}\le\theta\le \frac{\overline{Y}}{\frac32-\frac{9.8}{\sqrt{12n}}} $$
Note that $$\frac{1}{\frac32\pm\frac{9.8}{\sqrt{12n}}}\neq \frac23\pm\frac{\sqrt{12n}}{9.8}$$
To get confidence interval for $\theta^2$, simply square both ends of this interval. The only problem is that $\overline{Y}$ can be negative but for $n$ sufficiently large probabililty of this event is negligible since $\overline{Y}\to 1.5\theta >0$ in probabililty as $n\to\infty$.
Best Answer
It is correct. Find the 2 quantiles of the chi-squared assuming equiprobable tails and solve w.r.t. $\theta$
Your pivotal qty is not the only one possible.
Check the definition of $S^2$ because the one you posted is $=0$