Let $k$ be a field s.t. $f(x)$ degree $n\geq 3$ polynomial does not have repeated roots in algebraic closure. Consider $F(x,y)=y^2-f(x)$. Then $\operatorname{Spec}(k[x,y]/F)\subset P^2_k=\operatorname{Proj}(k[x_0,x_1,x_2])$ as scheme. WLOG, we can assume $x=\frac{x_1}{x_0},y=\frac{x_2}{x_0}$. Then homogenization gives $G(x_0,x_1,x_2)=x_0^{n-2}x_2^2-x_0^nf(\frac{x_1}{x_0})\in k[x_0,x_1,x_2]$.
Now one can use Jacobian criterion to tell whether $(0,0,1)$ point is singular or not. It is not singular when $n=3$. However, it will become singular once $n\geq 4$.
$\textbf{Q:}$ The book says "The one dimensional regular scheme without singular point is called hyperelliptic curve for $n\geq 5$ and for $n=4$, such an elliptic curve is actually an elliptic curve." Does this contradict with regularity checking by Jacobian criterion above? Above concludes the point at $(0,0,1)$ is always singular once $n\geq 4$.
Ref. Ueno, Algebraic Geometry 3, Chpt 7, Sec 2, Example 7.33