If you write your cubic as $y^2=f(x)$, let $M$ be the cardinality of of $S =\lbrace x \in \mathbb{F}_p, f(x)^{(p-1)/2} = 1 \rbrace$, so $M$ is related to the number of points on the cubic in an obvious way.
Define $G(x) = f(x)(f(x)^{(p-1)/2}-1) - f'(x)(x^p-x)/2$. Exercise, check that $G$ has double zeros on the elements of $S$. As $G$ has degree $3(p+1)/2$ we get $M \le 3(p+1)/4$ and $c_2=3/2, c_1=1/2$.
Edit: In the case of a general plane curve $f=0$ of degree $d$, you can use $G= (x^p-x)\partial f/\partial x + (y^p-y)\partial f/\partial y$. Again $G$ has double zeros on the $\mathbb{F}_p$-rational points of curve and meets the curve in finitely many points if $d$ is less than $p$ and $f=0$ has no linear component. So, in this case, the number of points is at most $d(d+p-1)/2$ by Bezout, i.e. $c_2= d/2$. Details are slightly harder to fill than the elliptic curve case. Also, there is no twisting so no corresponding lower bound.
