Given a simple closed curve $\gamma:[a,b]\subset\mathbb{R}\to\mathbb{R^2}, \gamma(t) = (x(t), y(t))$, $x(t)$ and $y(t)$ continuous in $\mathbb{R}$ and a point $P(x, y) = \gamma(t)$ for some $t \in [a,b]$, it is known that two normal unit vectors exist for $P$, which can be found by solving:
$ \begin{equation} \begin{cases} \mathbf{n}\cdot\gamma'(t)=0\\ ||\mathbf{n}||=1 \end{cases} \end{equation} $
Let the solutions be $\mathbf{n_1}$ and $\mathbf{n_2}$ with $\mathbf{n_1} = -\mathbf{n_2}$. Can one algebraically decide which of these unit vectors points outwards, i.e., without having to resort to plotting the curve and analyzing it?
Here's one way (among many), provided certain differentiability assumptions, though strictly speaking it involves calculus as well as algebra. One can ask if the velocity sweeps out a circle in the direction of the normal or in the opposite direction.
For the unit normal to be well defined and unique up to sign, the unit tangent vector $\mathbf{u}(t)=\dot{\gamma}(t)/\|\dot{\gamma}(t)\|$ has to exist everywhere. If $\mathbf{u}$ is differentiable, we can define the angular velocity function with respect to a particular choice of normal. $$ \omega(t)=\dot{\mathbf{u}}(t)\cdot\mathbf{n}(t) $$ It can then be shown (for a simple curve in $\mathbb{R}^2$) that $$ \int_a^b\omega(t)dt=\begin{cases}2\pi & \mathbf{n}\ \ \text{inward} \\ -2\pi & \mathbf{n} \ \ \text{outward} \end{cases} $$ Computing this quantity allows one to identify the two normals.