I need to study the continuity of this function:
$$ f(x, y) = \begin{cases} \frac{4y x^2}{x^4+ y^2}, & \text{if } (x, y) \neq (0, 0)\\ 0, & \text{if } (x, y) = (0, 0) \end{cases} $$
When I study the directional limits, specifically $y = px ^ 2$, I get the limit different than zero. But when I draw the function in 3D I see that the point (0.0) it is continuous.
Let us consider the limit of $f$ on a curve $y=px^2$. We can parametrize this curve by $t \mapsto (pt^2,t)$ and
$$\lim_{t\to 0} f(x(t),y(t)) = \lim \frac{4pt^4}{t^4+pt^4} = \frac{4p}{1+p}$$
So for $p=0$ the limit is actually $0$, but for $p=1$ we already get the limit $2$. Since these two limits are not equal, the function $f$ cannot be continuous at $0$.
Even if you plot this function at a low resolution in the domain $[-1,1]\times [-1,1]$, you can see that the limit on the curves $y=px^2$ is not zero for $p\neq 0$.