 Solution

For Newton's method for nonlinear systems, solve $J({\mathbf{z}}_0)\mathbf{\delta }=-\mathbf{F}({\mathbf{z}}_0)$ and set ${\mathbf{z}}_1={\mathbf{z}}_0+\mathbf{\delta }$. The Jacobian matrix here isAt $(0,0)$,The correction linear system is $3{\delta }_x-2{\delta }_y=1$, $-{\delta }_y=1$. Hence ${\delta }_y=-1$, ${\delta }_x=-1/3$, and one iteration givesAt the alternative starting value $(1,0)$,This Jacobian matrix is a singular matrix: its first row is twice its second. In fact the correction equations would bewhich are inconsistent, since twice the second left side equals the first while twice its right side is $0$, rather than $-3$. Thus Newton's iteration cannot take its first step from $(1,0)$. This is failure of the correction linear system, not merely nonuniqueness of a possible correction.