Confidence interval for exponential distribution with MLE

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 $\overline{Y}_n\dot{\sim}N\Big(\theta;\frac{\theta^2}{n}\Big)$

1. 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 $\chi_{(2n)}^2$