Let $q: \mathbb{R}^2 \to\mathbb{R}$ such that
$$ q(x,y):=\begin{cases}\left|\frac{y}{x^2}\right|\exp\left(-\left|\frac{y}{x^2}\right|\right),&\text{ for }x\neq 0\\0,&\text{ for }x=0\end{cases} $$
Can I say that for all $x$ and $y$ and for all $\epsilon>0$, there exists $\delta>0$ such that if $|q(x,y)−c|<δ$, then $|q(x,y)−q(c)|<ϵ$. So since $\exp$ is defined everywhere on $\Bbb R$, let $c \in \Bbb R$ and let $\epsilon>0$. But how do I find a corresponding $\delta>0$? Or is that already totally wrong?
Let us study the continuity of the function at the origin. For the function to be continuous there, the limit $$ \lim_{(x,y)\to (0,0)} \left|\frac{y}{x^2}\right|\exp\left(-\left|\frac{y}{x^2}\right|\right) $$ should exist and be equal to $q(0,0) = 0$. For the limit to exist, the function should approach the same value along any curve converging to the origin. However, if you consider the curve $(t,t^2)$, you see that, as $t\to 0$, $$ q(t, t^2) = \left|\frac{t^2}{t^2}\right|\exp\left(-\left|\frac{t^2}{t^2}\right|\right) = \exp(-1) \not\to 0. $$ Hence the function is discontinuous at the origin.