This is part of Exercise 7.3 in Undergraduate Algebraic Geometry by Reid.
Let $S: (f=0) \subset \mathbb{P}^3$ be a nonsingular cubic surface. For $P\in S$ prove that if $P$ is not on a line of $S$ then the intersection $S\cap T_P S$ is a cuspidal cubic if and only if the determinant of the Hessian matrix $h(f)(P)=0$. Deduce that cuspidal cubic sections exist, as required in the proof that there exists a line in the surface $S$.
I showed the first part using the Taylor expansion of $f$ restricted on the tangent plane at $P$, where I assumed $P=(1,0,0,0)$ and $T_PS: y=0$. Let the restriction be $\phi=f(1,0,z,t)$.
Then $\phi(P)=0$, and the linear part of $\phi$ at $P$ is zero. The quadratic part is either $cz^2$ or $ct^2$ if it is a cuspical cubic. So the determinant of Hessian matrix is zero.
Conversely, the determinant of Hessian matrix is zero implies that the quadratic part of $\phi$ is degenerate. If it is $czt$ then $f$ contains a term $xzt$. I am stuck here since this does not mean $P$ has to be on a line.
I found an example $f=z^3+t^3-xzt+x^2y$. In this case the restriction is $\phi=z^3-zt+t^3$, which is a nodal cubic. The determinant of $H(f)(P)=0$. And I couldn't see $P$ is on any line of $S$. Or maybe the line is not that obvious.
To show that cuspidal sections exist, I think I should use Bezout's theorem. There should be $12$ intersections of $f=0$ and $\det{h(f)}=0$. But again how to show that at least one of them is not on a line of $S$?
The problem is I don't know how to use the condition $P$ is not on a line of $S$.
Thank you for your help!
I finally figured out the converse part. I will write down my answer to close this question.
Suppose $H(f)(P)=0$ and $P$ is not on a line in $S$. Then $S\cap T_P(S)$ is an irreducible cubic curve $C$. Since $P$ is singular in $C$, $C$ can only be nodal: $xzt=t^3+z^3$, or cuspidal: $xz^2=t^3$.
An easy computation of $\phi$ finds out that only cuspidal cubic produces a degenerate quadratic part of $\phi$ which the zero Hessian determinant predicts. So the intersection must be cuspidal.
For the existence, by Bezout's theorem, $f=0$ and $H(f)=0$ has $12$ intersections. If the intersection point is not on a line, by the above argument, it generates a cuspidal cubic. If the intersection point is on a line, the cubic generated is reducible. In that case there exists a line in the surface $S$.