Question: Let $X\_1, X\_2 , \dots, X\_n$ be i.i.d observations from $N(\theta, \theta)$ (where $\theta$ is the variance). I intend to find the minimal sufficient statistics (M.S.S) for the joint distribution.

My steps ultimately lead me to the following:

$\dfrac{f\_\theta(x)}{f\_\theta(y)} = \text{exp}{ \left\{ \dfrac{-1}{2\theta} \left[ \left(\sum x\_i^2 - \sum y\_i^2 \right) + 2\theta \left( \sum y\_i - \sum x\_i \right) \right] \right) }$.

My M.S.S are therefore $\left( \sum x\_i^2 , \sum x\_i \right)$

Am I on track?

Answer: 
It turns out that after further expansion, the above becomes

$\dfrac{f\_\theta(x)}{f\_\theta(y)} = \exp \left\{ - \left[ \dfrac{1}{2\theta} \left( \sum x\_i^2 - \sum y\_i^2 \right) + \left( \sum y\_i - \sum x\_i \right) \right] \right\}$.

Since it is only the first term that depends on $\theta$, the M.S.S is $\sum x\_i^2$.
By writing the joint density in exponential family form we find

$$
f(x,\theta) = \frac1{\sqrt{2\pi}} \exp\left\{ -\frac{n}{2}\log\theta - \frac1{2\theta} \sum\_i x\_i^2 + \sum\_i x\_i -n\theta^2 \right\}
$$
and then the result follows by general exponential family theory, as noted in comments.