I know that in order for a set S to be smooth, it needs to be connected and that for each point $\vec{a} \in S$, there needs to be a neighborhood $N$ such that $S \cap N$ is a class $C^1$ function. The textbook I am reading from states if one is dealing with a parameterized function, it is usually a good idea to check if the function is one-to-one especially at points where the derivative is zero. For example, I am given the function $\vec{f}(t)=<\dfrac{t}{(t^2+1)},\dfrac{t}{(t^2+1)^2}>$ with $\dfrac{d\vec{f}(t)}{dt}=<\dfrac{1-t^2}{(t^2+1)^2},\dfrac{1-3t^2}{(t^2+1)^3}>$.
Here the points where the derivative are zero are $1,-1,\dfrac{1}{3}$ and $-\dfrac{1}{3}$. Clearly the function is one-to-one at these points and the curve only begins to approach a point where it might intersect itself is when $t \to \infty$ in which case the limit of f for both of its components are zero. Is it safe to say that the set S is not smooth?
For a parameterized curve, there will only be problems when all components of the derivative are 0. In this case, the first component is 0 when $t = \pm 1$ and the second component is zero as $t=\pm \frac{1}{\sqrt{3}}$. These don't overlap, so your example is smooth.
If you like, you may think of this like the following: Let $\vec{f}(t) = \langle x(t), y(t)\rangle$ represent a smooth curve. Suppose that $x'(t_0) \neq 0$. Then in a neighborhood of $t_0$, we can apply the inverse function theorem to write $t = t(x)$ as a function of $x$. This implies that $\vec{f}(t(x)) = \langle x, y(t(x))\rangle$, which is (locally) the graph of a function smooth function over the $x$-axis. Similarly, you can represent a portion of the curve as a smooth graph over the $y$-axis if $y'(t_0) \neq 0$.