I want to solve the quadratic equation $(\mathbf{A}\mathbf{x}).* \mathbf{x}=\mathbf{b}$ (I am using Matlab $.$ notation to denote the entrywise version of an operator) which means the following system of equations: $$ \begin{cases} \sum_{i=1}^n a_{1i} x_i x_1 = b_1, \\ \sum_{i=1}^n a_{2i} x_i x_2 = b_2, \\ \qquad \quad \vdots \\ \sum_{i=1}^n a_{ni} x_i x_n = b_n. \end{cases}$$
So far I have been simply iterating the function $\mathbf{f}(\mathbf{u})=\mathbf{b}./(\mathbf{A}\mathbf{u})$ (which means $\mathbf{v} = \mathbf{f}(\mathbf{u})$ is a vector such that $v_i = \frac{b_i}{\sum_{j=1}^n a_{ij} u_j}$). When does this method converge on an answer? What are some conditions ensuring the existence or uniqueness of solutions?
I agree with @Carl Christian, about the next step of your iteration.
I propose you a result in the case where the underlying space is $\mathbb{R^2}$:
In general, there exist at most 4 solutions. This is a consequence of a graphical understanding. In fact;
$(Ax)\circ x=b$ with $A=\begin{pmatrix} p & r \\ q & s \end{pmatrix}$ gives
$(px+ry)x=b_1$ and
$(qx+sy)y=b_2$,
The problem's solutions are (if they exist) the common points to two conical curves, which have in general at most 4 common points.
These conics are hyperbolas, both centered at the origin, the first (resp. second) one with the $y$ axis (resp $x$-axis) as one of its asymptotes. It looks as if in fact there could be at most 2 common points in the general case.
Some degeneracy may occur. An example : if rank(A)=1 and $b_1=b_2=0$, all the points of the line with equation $px+ry=0$ are solutions.
You can find these solutions by applying the 2D (iterative) Newton method to the above system.