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