Suppose that $\phi$ was the characteristic function of a 2d random vector $X=(X_1,X_2)$. Then, $X_1$ and $X_2$ have the characteristic functions $\phi(t,0)=\exp(-t^2+it/3)$ and $\phi(0,t)=\exp(-t^2)$ respectively. This means that $X_1$ and $X_2$ are each normally distributed (but not joint normal). Then, $ \begin{align} \exp(t\Vert X\Vert)&\le\exp(t\vert X_1\vert)\exp(t\vert X_2\vert)\\ &\le\max\left(e^{t\vert X_1\vert},e^{t\vert X_2\vert}\right)^2\\ &\le e^{2t\vert X_1\vert}+e^{2t\vert X_2\vert} \end{align} $ has finite mean for all positive $t$. This means that $\phi(t_1,t_2)=\mathbb{E}[\exp(it_1X_1+it_2X_2)]$ extends to an analytic function of $t_1,t_2$ everywhere on $\mathbb{C}^2$. By uniqueness of analytic continuation, this must agree with your expression on $\mathbb{C}^2$. However, your expression does not converge as $t_2\to it_1$.
I'll add another proof that $\phi$ is not a characteristic function, which is more closely related to the title of the paper referenced in the question (Non-stable laws with all projections stable). If $\phi$ was the characteristic function of a random vector $X=(X_1,X_2)$ then, for any $a\in\mathbb{R}^2$, $a\cdot X$ would have characteristic function $ \mathbb{E}\left[e^{ita\cdot X}\right]=\phi(ta_1,ta_2)=\exp\left(-\Vert a\Vert^2 t^2+it\frac{a_1}{3}\frac{a_1^2-3a_2^2}{a_1^2+a_2^2}\right). $ So, $a\cdot X$ is normally distributed with variance $2\Vert a\Vert^2$ and mean $\mu(a)\equiv\frac{a_1}{3}\frac{a_1^2-3a_2^2}{a_1^2+a_2^2}$. This is contradictory, as it does not even respect additivity of expectations, $ \mathbb{E}[(a+b)\cdot X]=\mu(a+b)\not=\mu(a)+\mu(b)=\mathbb{E}[a\cdot X]+\mathbb{E}[b\cdot X]. $