Answer by WHLin for Why does $f(x,y)= \frac{xy^2}{x^2+y^4}$ have different limits when approaching $(0,0)$ along straight lines vs. along the curve $(1/t^2,1/t)$?

This has something to do with the uniform continuity around 0. Given a fixed $\epsilon > 0$, if there exist an universal constant $\delta_u$ such that

$$|(x,y)|<\delta_u \implies |f(x,y)|<\epsilon$$

then your intuition is right: along any curve go to 0, $f$ will have limit 0.

However, for this example the origin is not uniformly continuous. Along lines $y=ax$, $a \in\mathbb{R}$ we can define a class of fucntions $f_a(r) = \frac{a^2 x}{1+a^4 x^2}$. Suppose the universal constant $\delta_u$ exist, then for every $a$ we must have

$$ |x|<\delta_u \implies |f_a(x)|<\epsilon $$

However, let $a=\frac{1}{\sqrt{\delta_u}}$ we have $f_a(\delta_u) = 1/2$. Therefore, above argument cannot be correct, and the origin is not uniform continuous with respect to $\{f_a(x)\}$.