If the limit of a multivariable function is identical along a certain class of paths, can we claim the existence of the limit?

EDITED (Final version).

This example shows that no reasonably simple class of paths suffices to analyze the behavior at a point of a given function $f\colon \mathbb{R}^2\to \mathbb{R}$. Let $\epsilon\colon \mathbb{R}\to \mathbb{R}$ be such that $\epsilon(y)\to 0$ as $y\to 0$ and define \begin{equation} f(x, y)=\begin{cases} \displaystyle \frac{\left( \epsilon(y) - x\right)^2}{\epsilon(y)^2 + x^2} , & (x, y)\ne (0,0) \\ \displaystyle 1, & (x, y)=(0,0) \end{cases} \end{equation} Any substitution $$\begin{array}{ccccc} x=\gamma(y),& \text{with }&\epsilon=o(\gamma)&\text{or}& \gamma=o(\epsilon)\end{array} $$ produces the limit $$ \lim_{y\to 0} f(\gamma(y), y)= 1, $$ so the function is continuous along the path. However, the substitution $$ x=m\epsilon(y),\qquad m \in\mathbb{R} $$ produces the limit $$ \lim_{y\to 0} f(m\epsilon(y), y)=1-\frac{2m}{1+m^2}, $$ so the function is almost never continuous along this path (this only happens in the case $m=0$).

Choosing $\epsilon(y)=\exp\left(-y^{-2}\right)$ produces a discontinuous function $f=f(x, y)$ that is continuous along all polynomial paths $$ x=a_1y + a_2 y^2 +\dots + a_ny^n,\qquad n\in\mathbb{N}. $$

This said, the question does admit a positive answer, but of little practical use I am afraid:

Fact. Let $f\colon \mathbb{R}^2\to \mathbb{R}$. The following are equivalent:

1. One has that $\displaystyle \lim_{(x, y)\to (0,0)} f(x, y)=L$.

2. For any sequence $(x_n, y_n)\to (0,0)$, one has that $f(x_n, y_n)\to L$.

3. For any continuous curve $\gamma\colon [0, 1]\to\mathbb{R}^2$ such that $\gamma(0)=(0,0)$ one has that $\displaystyle \lim_{t\to 0} f(\gamma(t))=L$.

4. For any smooth curve $\eta \colon [0,1]\to\mathbb{R}^2$ such that $\eta(0)=(0,0)$ one has that $\displaystyle \lim_{t \to 0} f(\eta(t))=L$.

proof. It is clear that $1.\Leftrightarrow 2. \Rightarrow 3.\Rightarrow 4.$ Let us only prove the converse implication $3.\Rightarrow 2.$ Consider a sequence $(x_n, y_n)\to (0,0)$. Define a continuous path $\gamma\colon [0,1]\to \mathbb{R}^2$ as follows: $$ \gamma(t) = \Big( (n+1)(nt-1)x_{n+1} + n[(n+1)t-1]x_n ; (n+1)(nt-1)y_{n+1} + n[(n+1)t -1]y_n\Big)$$ for $t\in\left[\frac{1}{n+1}, \frac1n\right]$. (This is the piecewise linear path with the property that $\gamma\left(\frac1n\right)=(x_n, y_n)$ and $\gamma\left(\frac{1}{n+1}\right)=(x_{n+1}, y_{n+1})$). By assumption, $$ f(\gamma(t))\to L,\qquad t\to 0.$$ Therefore, $$ f(x_n, y_n)=f\left(\gamma\left(\frac1n\right)\right) \to L,\qquad n\to\infty.$$ $\square$

Remark. The proof that $4.\Rightarrow 2.$ is the same, with the only added technical difficulty that the path $\gamma$ must be constructed smooth.