An elliptic curve $E$ over a field $K$ is a projective smooth curve of genus $1$ defined over $K$ with at least one $K$-rational point. This means that there is a non-singular model for $E$, but this does not mean that every curve that is birationally equivalent to $E$ is necessarily non-singular everywhere.
For example, take $E:y^2=x^3+1$ and consider $C: X^2Y^2 = X^3+1$. The curves $E$ and $C$ are birationally equivalent via $\phi: E\to C$ such that $\phi(x,y)=(x,x/y)$. Clearly, $\phi$ is such an equivalence, with inverse $\phi^{-1}: C\to E$ given by $\phi^{-1}(X,Y)=(X,XY)$. However, the curve $E$ is non-singular, but $C$ is not. The curve $C$ is singular at $[0,1,0]$, because in projective coordinates $C: F(X,Y,Z)=0$ with $F(X,Y,Z)=X^2Y^2-X^3Z-Z^4$ and
$$\left(\frac{\partial F}{\partial X},\frac{\partial F}{\partial Y},\frac{\partial F}{\partial Z}\right)=(2XY^2-3X^2Z,2X^2Y,-X^3-4Z^3),$$
which vanishes at $[0,1,0]\in C$.
I believe Magma is simply saying that your curve $C$ is genus $1$ and has a non-singular model $E$.
Short answer: the point at infinity is an easy case. For any affine point, notice that $\det H=8\psi_3$, where $\psi_3$ is a division polynomial. From standard properties of division polynomials, the result follows.
Longer answer: in case you don't immediately see the connection with the division polynomial, you could take the following approach. This is what I started out with, but essentially what we're doing is deriving $\psi_3$.
Let's first assume that $P\neq \mathcal{O}$ and $[2]P\neq\mathcal{O}$, in those cases we would want to show that the Hessian always respectively never has determinant zero. This is done at the end. Write $P=(x,y)$.
In this case we know that the tangent line contains a third point, and it is not $\mathcal{O}$. In fact, we have exact formulas to compute the third point (see page 54 of Silverman, specialized to $\operatorname{char}\neq2,3$), namely:
$$\begin{align*}
\lambda &= \frac{3x^2+A}{2y}, \\
x_{2} &= \lambda^2-2x, \\
y_{2} &= \lambda (x-x_2)-y.
\end{align*}$$
Note that $(x_2,y_2)$ is generally derived by taking it as the third intersection point of the tangent line at $P$. Now we know that $[3]P=\mathcal{O}\iff x_2=x$ and $y_2=-y$. Substituting the curve equation into $x=\lambda^2-2x$, we obtain $$3x^4+6Ax^2+12Bx-A^2=0.\quad (\text{this is }\psi_3\text{!})$$
Now I will not post the whole computation here (I will if you really want me to!), but as it turns out the left-hand side is almost the determinant of the Hessian. So we can conclude that if $[3]P=\mathcal{O}$, then the determinant of the Hessian vanishes.
On the other hand, if the determinant vanishes, we can revert all operations above to show that $x=\lambda^2-2x$. Hence $x_2=x$. From this it immediately follows that $y_2=-y$, hence $[2]P=-P$, and $[3]P=\mathcal{O}$.
Finally a note on $P=\mathcal{O}$ and $[2]P=\mathcal{O}$. In the first case we write $P=(0:1:0)$ and it is immediate from a computation that the Hessian has zero determinant. If $[2]P=\mathcal{O}$, then we know that $y=0$. We can compute the determinant as $D(x)=24Ax^2+72Bx-8A^2$. As $y=0$, also $F(x)=x^3+Ax+B=0$. It turns out that $$\operatorname{Res}(D,F)=-\Delta^2,$$
where $\Delta$ is the (non-zero!) discriminant of the curve. Hence $D$ and $F$ do not share a root, and we conclude that $D(x)\neq 0$.
Best Answer
To prove this I'm going to appeal to some results in algebraic geometry. We have that $P$ is a simple point on $F$ if and only if $\mathcal{O}_P(F)$, the local ring of $F$ at $P$, is a discrete valuation ring (see Fulton theorem 1 section 3.2. A simple point $P$ on $F$ is called an ordinary flex if $ord_p^F(L) = 3$ where $L$ is the tangent line to $F$ at $P$.
By a theorem in section 5.3 (again see Fulton), $I(P, H \cap F) = 1$ if and only if $P$ is an ordinary flex. Since an elliptic curve is nonsingular, all of its points are simple. Since we are looking for the number of $3$-torsion points, these are exactly the points that are ordinary flexes. By Bezout's theorem we have that, since the Hessian is a cubic polynomial and elliptic curves have degree $3$, $$ \sum_P I(P, H \cap F) = 9 $$ so $F$ has at most nine ordinary flexes.