Assume $f:\mathbb R^2\to\mathbb R$ is such that for all $\varepsilon>0$ exists $\delta>0$ such that, whenever $||x||<\delta$, also $||f(x)||<\varepsilon^2$. Can we see that $f$ is $\Theta(||x||)$?
In other words, can we see that $$\lim_{x_1\to 0}_{x_2\to 0}\frac{f(x_1,x_2)}{\sqrt{x_1^2+x_2^2}}=0$$
What are sufficient conditions for $f$ to be $\Theta(||x||)$?
My problem stems here.
No, we cannot. Note, that $f(x) = \|x\|$ fulfills your definition, given $\epsilon>0$, choose $\delta = \epsilon^2$.