We say that a curve $F(x,y)=0$ is regular at $(x_0,y_0)$ if in a nbd of that point,the curve can be written explicitly as a continuously differentiable function of the form $y=f(x)$ or $x=g(y)$.We know by implicit function theorem,we know if$F^2_x+F^2_y\neq0$,at a point,then the curve represents an implicit function $f$ such that $(x,f(x))$ satisfies $F(x,y)=0$ or $(g(y),y)$ satisfies $F(x,y)=0$.Now my question is can an algebraic curve be regular even if $F^2_x+F^2_y=0$?
Also if we consider the curve $F(x,y)=0$ where $F(x,y)=y^2-x^6$,then $\partial F/\partial x |_{(0,0)}=\partial F/\partial y|_{(0,0)}=0$But the curve has a tangent at $(0,0)$ which is the $x$-axis although $dy/dx$ makes no sense as $dy/dx=$$-$${\partial F/\partial x}\over{\partial F/\partial y}$ has $0$ in denominator.
Also consider the cusp of $y^3-x^2=0$,it is described as singularity in Courant although $y$ is expressible as $ ^3\sqrt {x^2}$,so it should be regular as per definition but it is described as a singular point.
As you see continuously differentiable on a real interval doesn't seem to be a good definition.
The continuously differentiable assumption works better for complex curves because for algebraic functions continuously differentiable implies holomorphic which implies analytic. Neither $x^{3/2}$ nor its "derivative" $x^{1/2}$ are continuous near $0\in \Bbb{C}$ ( $e^{i \theta 3/2}r^{3/2},\theta \in [-\pi/2,\pi /2)$ has a branch cut at $\theta=\pi/2$).
For such algebraic functions it suffices to assume it is smooth on a real interval to get that it is holomorphic on a complex neighborhood of it.