Let Y1,...Yn be a random sample from $N(Mean = \theta,Var = \theta)$
Show $U = \frac{(n-1)S^2}{\theta^2} $ is a pivotal quantity.
$ S^2 = \sum_{i=1}^{n} (Y_i - \overline{Y})/(n-1) $
so $U = \frac{(n-1)S^2}{\theta} $ is a chi-square but what about over $\theta^2$?
What distribution is U?
Remember that if $X_1,\ldots X_n$ are a sample from the normal($\mu,\sigma^2$) distribution, then $\sum(X_i-\bar X)^2/\sigma^2$ has chi-square distribution with $n-1$ degrees of freedom.
To apply this to your problem, define $$Z_i:=\frac{Y_i-\theta}\theta.$$ Argue that each $Z_i$ has standard normal distribution. Next, use algebra to show that $$\sum(Y_i-\bar Y)^2=\theta^2\sum(Z_i-\bar Z)^2. $$ From this you should be able to conclude that $$U:=\frac{(n-1)S^2}{\theta^2}=\frac{\sum(Y_i-\bar Y)^2}{\theta^2}$$ has chi-square distribution with $n-1$ df, which is free of $\theta$, and is therefore pivotal.