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Let $Y_1, ... Y_n $ be a random sample from $N(\theta,\theta) $

use $U = \sqrt{n}(\overline{Y}-\theta)/\sqrt{\theta}$ to derive a $(1-\alpha)$ confidence interval for $\theta$

since U is normal $P[-Z_{\alpha/2} < U < Z_{\alpha/2}] = 1-\alpha $ and try to solve for $\theta$ but there is no way to separate $\theta$

So how do I obtain an inequality for $\theta$ ?

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    With a normal distribution, one should always specify whether 2nd argument in the notation is SD or variance. I suspect variance here and it appears so did @NCh. – 2017-02-23

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You can solve two quadratic inequalities and find confidence interval $$ \frac12\left(-\frac{Z_{\alpha/2}}{\sqrt n}+\sqrt{\frac{Z^2_{\alpha/2}}{n}+4{\overline Y}}\right)\leq \theta\leq \frac12\left(\frac{Z_{\alpha/2}}{\sqrt n}+\sqrt{\frac{Z^2_{\alpha/2}}{n}+4{\overline Y}}\right) $$ Unfortunately, this interval is valid only for the case when $\frac{Z^2_{\alpha/2}}{n}+4{\overline Y}\geq 0$.

For the opposite case the inequality $-Z_{\alpha/2} < U < Z_{\alpha/2}$ is impossible and has none solutions. Fortunately, $\overline Y\xrightarrow{p}\theta>0$ and $P(\overline Y >0)\to 1$ as $n\to\infty$.