 Solution

For $V=Z(f_1,\dots ,f_r)$, $T_{V,p}$ is the common kernel of $d{f}_i(p)$; $\dim V$ is the transcendence degree of its function field, and $p$ is smooth when $\dim T_{V,p}=\dim V$. A nonzero maximal Jacobian minor defines a nonempty open smooth locus (generic rank gives nonemptiness).

A line $ux+vy+wz=0$ corresponds to $[u:v:w]\in ({\mathbb{P}}^2{)}^{\ast }$. Lines through $P=[a:b:c]$ satisfy $au+bv+cw=0$, a projective line. For a smooth curve $F=0$, the dual morphism is $P\mapsto [F_x(P):F_y(P):F_z(P)]$.

For $F=x^3+y^3+z^3$, eliminating $x,y,z$ from $[u:v:w]=[x^2:y^2:z^2]$ givesThe dual sextic is singular: its nine cusps are the tangent lines at the nine flexes of the cubic. Thus points of $X$ mapping to singularities are precisely inflection points, where the tangent has higher contact.