MLE of $\theta$ in $N(\theta, \theta^2)$ 
I have $X_i \overset{iid}{\sim} N(\theta, \theta^2)$, $\theta >0$, $i=1, \cdots, n$. I would like to find the MLE for $\theta$.

My attempt
I have a log likelihood $l(\theta) = -\frac{n}{2} \log(2 \pi) - \frac{\sum x_i^2}{2 \theta^2}  + \frac{\sum x_i}{\theta} - \frac{n}{2}$. 
$$
\begin{aligned}
\frac{\partial}{\partial \theta} l(\theta) &= \sum x_i^2/\theta^3 - \sum x_i/\theta^2 \\
\frac{\partial^2}{\partial \theta^2} l(\theta) &= -3\sum x_i^2/\theta^4 + 2 \sum x_i/\theta^3 \\
&= \frac{-3 \sum x_i^2 + 2 \sum x_i \theta} {\theta^4}
\end{aligned}
$$
From $\sum x_i^2/n > \left( \sum x_i \right/n)^2$, I have when $0 < \theta < \frac{3}{2n} \sum x_i$, $\frac{\partial^2}{\partial \theta^2} l(\theta) < 0$
So by setting $\frac{\partial}{\partial \theta} l(\theta) = 0$, I have $\hat{\theta}^{MLE} = \sum x_i^2 / {\sum x_i}$ when $\sum x_i >0 $,
But I can't proceed with the cases $\sum x_i < 0$ or $\frac{3}{2n} \sum x_i < \theta$. 
 A: That calculation doesn't look quite right.
Given the sample $(x_1,\ldots,x_n)$, likelihood function of $\theta$ is
\begin{align}
L(\theta\mid x_1,\ldots,x_n)&=\frac{1}{(\theta\sqrt{2\pi})^n}\exp\left[-\frac{1}{2\theta^2}\sum_{i=1}^n(x_i-\theta)^2\right]\quad,\theta>0
\end{align}
So the log-likelihood is of the form
$$\ell(\theta)=\text{constant}-n\ln\theta-\frac{1}{2\theta^2}\sum_{i=1}^n(x_i-\theta)^2$$
, so that
$$\frac{\partial\ell}{\partial\theta}=\frac{-n}{\theta}-\frac{n\bar x}{\theta^2}+\frac{\sum_{i=1}^nx_i^2}{\theta^3}$$
Setting $\dfrac{\partial\ell}{\partial\theta}=0$ and ignoring negative value of $\theta$ yields
$$\hat\theta=-\frac{\bar X}{2}+\sqrt{\frac{\sum_{i=1}^n X_i^2}{n}+\frac{\bar X^2}{4}}$$
Verify that this is indeed the MLE by checking $$\frac{\partial^2\ell}{\partial\theta^2}\mid_{\theta=\hat\theta}<0$$
Finally we should arrive at
$$\hat\theta_{\text{MLE}}=-\frac{\bar X}{2}+\sqrt{\frac{\sum_{i=1}^n X_i^2}{n}+\frac{\bar X^2}{4}}\quad,\text{ if }\bar X>0$$
A: Since $\theta >0$ it suffices to show the numerator is always negative. Note that you only need to show it's negative at the MLE, not all values of $\theta$.
$$-3\sum x_i^2 + 2 \sum x_i \theta <0$$
Substitute in the MLE to obtain
$$-3 \sum_i x_i^2 + 2\sum_i x_i \frac{\sum_j x_j^2}{\sum_j x_j}$$
Cancel off the two sums on the right hand term
$$=-3\sum_i x_i^2 + 2 \sum_j x_j^2$$
$$=-1 \sum_i x_i^2 <0$$
Which is negative since every summand $x_i^2 >0$
Hence since the second derivative is negative at $\theta^{MLE}$, it's a local maximum of the likelihood function.

As pointed out by @StubbonAtom your derived MLE is incorrect. For example, examine the following R code with $\theta = 2$, the estimator you derived gives $\hat{\theta} \approx 4$.
x<- rnorm(10000,2,2)
sum(x^2)/sum(x)
[1] 4.009042

