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 :/
How can we given a approximately confidence interval...
This is done with the asymptotic distribution of $\hat{\theta}_{ML}$ that is Gaussian
Your (correct) attempt leads to the exact confidence interval
What do you wanna do?
Asymptotic CI
$\hat{\theta}=\overline{Y}_n$
It is known that $\overline{Y}_n\dot{\sim}N\Big(\theta;\frac{\theta^2}{n}\Big)$
- First way to proceed
Estimate the standard deviation of $\overline{Y}_n$ with $\frac{\overline{Y}_n}{\sqrt{n}}$ and solve the following double inequality in $\theta$
$$-z < \frac{\overline{Y}_n-\theta}{\overline{Y}_n}\sqrt{n}<z$$
2. Second method
Solve in $\theta$ the following double inequality
$$-z < \frac{\overline{Y}_n-\theta}{\theta}\sqrt{n}<z$$
I prefer the third method, the exact one your posted in your attempt (is easier, related with a $\chi^2$
In fact, as you showed $\Sigma_i Y_i\sim Gamma(n;\frac{1}{\theta})$ thus
$\frac{2n}{\theta}\overline{Y}_n\sim Gamma(n;\frac{1}{2})=\chi_{(2n)}^2$
Thus you can solve the following doubel inequality in $\theta$
$$a<\frac{2n}{\theta}\overline{Y}_n<b$$
where $a,b$ are the quantiles of the $\chi_{(2n)}^2$
Best Answer
Multiply both parts by sum of squares and then multiply by two: $$ \mathbb P\left(\theta<a\frac{n}{\sum_{i=1}^n x_i^2}\right)=\mathbb P\left(\theta \sum_{i=1}^n x_i^2 < an\right) = \mathbb P\left(2\theta \sum_{i=1}^n x_i^2 < 2an\right)=0.1 $$ Note that the l.h.s. $2\theta \sum_{i=1}^n x_i^2$ has Gamma distribution $\Gamma\bigl(n,\frac12\bigr)=\chi^2_{2n}$ since $2\theta x_1^2$ has an exponential distribution with parameter $\frac12$. Next you can use $0.1$-quantile of chi-square distribution with $2n$ degrees of freedom: $$ \mathbb P(\chi^2_{2n}<h_{0.1})=0.1 $$ and take $2an=h_{0.1}$, $a=\frac{h_{0.1}}{2n}$.
For $b$, $0.9$-quantile of chi-square distribution with $2n$ degrees of freedom is needed: $$ \mathbb P(\chi^2_{2n}>h_{0.9})=0.1 $$ and take $2bn=h_{0.9}$, $b=\frac{h_{0.9}}{2n}$.
Finally, confidence interval will be $$\left(\frac{h_{0.1}}{2n}T,\,\frac{h_{0.9}}{2n}T\right)=\left(\frac{h_{0.1}}{2\sum_{i=1}^n x_i^2},\,\frac{h_{0.9}}{2\sum_{i=1}^n x_i^2}\right)$$