Let $F$ be a square-free (as a polynomial) ternary cubic form over $\mathbb{C}$, and let $H_F$ be its Hessian determinant... which is also a ternary cbic form. If $F$ splits over $\mathbb{C}$, so that it can be written as the product of three linearly independent linear forms (i.e., the $3 \times 3$ matrix formed by the coefficients of the three linear forms is non-singular), then one has that $F, H_F$ are proportional. Indeed, since $H_F$ is a covariant and any such $F$ is $\text{GL}_3(\mathbb{C})$-equivalent to $xyz$, it suffices to check this assertion for this form, and this is trivial.

The converse is not true. For example, for $F = x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - xz - yz)$ we also have $F, H_F$ are proportional. Moreover, since the quadratic factor is equal to $((x-y)^2 + (x-z)^2 + (y-z)^2)/2$ which is positive definite and non-singular, it cannot be a product of linear forms over $\mathbb{C}$.

Thus, a necessary condition for $F$ to split into linear factors over $\mathbb{C}$ is for $F, H_F$ to be proportional, but this is not sufficient by the example above. What additional condition is needed to ensure that $F$ is in fact a product of linear forms?

Edit: it seems I was too quick to claim that $(x-z)^2 + (x-y)^2 + (y-z)^2$ is *positive* definite. In fact it has the non-trivial solution $x = y = z = 1$. Moreover, the matrix $\left(\begin{smallmatrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 0 & 1 & -1 \end{smallmatrix}\right)$ is singular. As Will Jagy points out below, it factors as $(x + \omega y + \omega^2 z)(x + \omega^2 y + \omega z)$ for $\omega$ a primitive third root of unity.