In Definition 3.4.7 of Statistical Inference by Casella & Berger, 2nd ed., a curved exponential family is defined to be a family of the form
$$ f(x \vert \boldsymbol\theta) = h(x)c(\boldsymbol\theta)\exp\left(\sum_{i=1}^{k}w_i(\boldsymbol\theta)t_i(x)\right) $$
for which the dimension of the vector $\boldsymbol\theta$ is equal to $d<k$. A normal $n(\theta, \theta)$ (i.e. mean and variance are both $\theta$) is used in problem 3.33 and in Example 3.4.9 as an example of a curved exponential family. But I'm having trouble seeing why this is a curved family. Since there is only one parameter, $d=1$. Working out the form of the exponential family to which this belongs, I get
$$ \frac{1}{\sqrt{2\pi\theta}}\exp\left(\frac{\theta}{2}\right)\exp(x)\exp\left(-\frac{x^2}{2\theta}\right) $$
so that $h(x) = \exp(x)$, $c(\theta) = \frac{\exp\left(\frac{\theta}{2}\right)}{\sqrt{2\pi\theta}}$, and $w_1(\theta) = -1/2\theta$, $t_1(x) = x^2$. In that case, $k=1$. So we have a situation where $d=k$, which does not meet the condition for a curved exponential family according to the above definition. I'm confused about why the text considers it a curved family in Example 3.4.9 and Exercise 3.33(a). Can anyone clarify this for me?
Exercise 3.33 doesn't ask you to verify that the family is a curved exponential family, it only asks you to verify that the family is exponential. You are correct in that your calculation shows that $N(\theta, \theta)$ is a full exponential family (the second part of Definition 3.4.7). Here's a screenshot of the exercise in question.
By asking you to describe the curve on which the parameter vector lies, the authors mean the ray $(\theta, \theta)$ for all $\theta > 0$.
A second, minor point is that you've also dropped a sign in your derivation, which follows.
$$f(x|\theta) = \frac{1}{\sqrt{2 \pi \theta}} \exp\left( - \frac{(x-\theta)^2}{2 \theta}\right)$$ $$ = \frac{1}{\sqrt{2 \pi \theta} } \exp \left(-\frac{x^2}{2 \theta} + x + -\frac{\theta}{2} \right)$$ $$ = \frac{1}{\sqrt{2 \pi \theta} } \exp(x) \exp(-\frac{\theta}{2}) \exp \left(-\frac{x^2}{2 \theta} \right).$$
UPDATE: The OPs question was more subtle than I gave it credit for. See the comments below for details.