I would like to see some mention of the fact that the division by the gradient is a meaningless, purely formal motivation for the correct step (inverting the Hessian) that follows.
f(x) =~ f(a) + f'(a)(x-a)
0 = f(a) + f'(a)(x-a)
0 = f'(a)^-1 f(a) + x-a
x = a - f'(a)^-1 f(a)
If f is a multi dimensional function f : R^n -> R^n then the derivative f'(a) is the Jacobian, and inversion becomes matrix inversion.
When we use Newton's method for minimisation of a function g we solve g'(x) = 0, so we pick f(x) = g'(x). Since the formula above contains f' we get a second derivative. The second derivative in multiple dimensions is the hessian.
If you aren't using LaTeX for formatting, then think partial^2 of l. FWIW: I just found this[2] which would make (using the physics package) even simpler to represent.
[1] https://en.wikipedia.org/wiki/Hessian_matrix
[2] https://tex.stackexchange.com/questions/225523/how-to-write-...
As to the motivation for the correct step: can you point me to a resource that explains this? Not sure I follow...
You write an equation involving division by the gradient. This is an illegal operation (one cannot divide by a vector), and your final recipe doesn't do it. As far as I can tell, you are writing down the incorrect, illegally-vector-inverting formula as motivation for the correct formula involving the (inverse of the) Hessian. All I am suggesting is that you say explicitly something like "Of course, this formula as written is not literally correct; one cannot actually divide by a vector. The correct procedure is explained below."
(Incidentally, speaking of inverses, another poster (https://news.ycombinator.com/item?id=14881265) has mentioned that it may be a bit confusing to speak of the inverse of a matrix rather than the reciprocal, since (as I interpret that other poster's point) the reciprocal of a matrix is just its inverse. I might prefer to say something like "We write $H_{\ell(\theta)}^{-1}\nabla\ell(\theta)$ rather than $\frac{\nabla\ell(\theta)}{H_\ell(\theta)}$ to emphasise that we are inverting a matrix, not a scalar, so that the order of multiplication matters.")