Show $\frac{(n-1)S^2}{\theta}$ is pivotal quantity of random sample $Y_1,...,Y_n$ from $N(\theta,\theta)$ then derive confidence interval 
Show $\frac{(n-1)S^2}{\theta}$ is pivotal quantity of random sample $Y_1,...,Y_n$ from $N(\theta,\theta)$ then derive confidence interval $(1-\alpha)100%$ confidence interval for $\theta$

So I can show that $\frac{(n-1)S^2}{\theta}\sim \chi^2_{n-1}$ since $S^2=\sum_{k=1}^n \frac{(Y_i-\bar{Y})}{n-1}$
But I'm not sure how to find the confidence interval?
I believe it is supposed to look like $P(\chi^2_{n-1}\leq \frac{(n-1)S^2}{\theta}\leq \chi^2_{n-1})=1-\alpha$
And then solving for $\theta$ in the middle. But I know $\chi^2_{n-1}$ is wrong, I don't understand what the bounds are supposed to be.
 A: 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 $=0$
A: Suppose $\theta = \mu = \sigma^2,$ data are a random sample for a normal distribution with these parameters,
$S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i = \bar X)^2,$ and the pivotal quantity $Q = \frac{(n-1)S^2}{\theta} \sim \mathsf{Chisq}(\nu = n-1).$
Then choose $L, U$ to cut probability $0.025$ from the lower and upper tails, respectively, of $\mathsf{Chisq}(n-1).$
$$P\left(L < \frac{(n-1)S^2}{\theta} <R\right)
= P\left(\frac{(n-1)S^2}{U}<\theta<\frac{(n-1)S^2}{L}\right)= 0.95.$$
Thus a 95% confidence interval for $\theta$ is of the form
$\left(\frac{(n-1)S^2}{U},\,\frac{(n-1)S^2}{L}\right).$
For example, if $n = 10, \theta = 7$ then data might be as simulated in R below:
set.seed(2021)
x = rnorm(10, 7, sqrt(7))
summary(x);  var(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  1.913   7.200   7.937   7.803   9.147  11.577 
[1] 6.263132  # sample variance

And the 95% CI for $\theta$ is $(2.96, 20.97),$
9*var(x)/qchisq(c(.975,.025), 9)
[1]  2.963196 20.874096

Note: It seems worthwhile looking at this a little more extensively. The 95% t interval $(6.01,9.59)$ for $\theta$ is noticeably shorter than the CI you propose:
set.seed(2021)
x = rnorm(10, 7, sqrt(7))
t.test(x)$conf.int
[1] 6.012665 9.593205
attr(,"conf.level")
[1] 0.95

