how to find the zero root of only one equation with two arguments by Newton-Raphson's method? The two dimensional Newton-Raphson method is to find a zero root $(x_0,y_0)$ which satisfy
\begin{array}{*{20}{c}}
  {f\left( {{x_0},{y_0}} \right) = 0} \\ 
  {g\left( {{x_0},{y_0}} \right) = 0} 
\end{array} by iteration.
However, now I have only one equation $f(x,y)=0$ and I want to find its zero root $(x_0,y_0)$, what improvement should I do to the the Newton-Raphson method?
If I add another equation $g(x,y)=0$, then the determinant of Jacobian matrix will be zero and the inverting of the Jacobian matrix wouldn't exist.
 A: $$
{\rm f}\left(\vec{r}\right)
+
\delta\vec{r}\cdot\nabla{\rm f}\left(\vec{r}\right) \approx 0
$$
$$
{\rm f} + \left(\delta x\quad\delta y\right){{\rm f}_{x} \choose {\rm f}_{y}} \approx
0
$$
$$
{\rm f}\left({\rm f}_{x}\quad{\rm f}_{y}\right)
+
\left(\delta x\quad\delta y\right)
\left({\rm f}_{x}^{2} + {\rm f}_{y}^{2}\right) \approx
0
$$
$$
\left(\delta x\quad\delta y\right)
\approx
-\,
{{\rm f}\left({\rm f}_{x}\quad{\rm f}_{y}\right)
 \over
\left({\rm f}_{x}^{2} + {\rm f}_{y}^{2}\right)}
$$
$$
\left\lbrace%
\begin{array}{rcl}
\delta x
& \approx &
-\,
{{\rm f}\left(\vec{r}\right)
 \over
 \left\lbrack{\partial{\rm f}\left(\vec{r}\right)\over\partial x}\right\rbrack^{2}
 +
 \left\lbrack{\partial{\rm f}\left(\vec{r}\right)\over\partial y}\right\rbrack^{2}
}\,
{\partial{\rm f}\left(\vec{r}\right)\over\partial x}
\\[3mm]
\delta y
& \approx &
-\,
{{\rm f}\left(\vec{r}\right)\,
 \over
 \left\lbrack{\partial{\rm f}\left(\vec{r}\right)\over\partial x}\right\rbrack^{2}
 +
 \left\lbrack{\partial{\rm f}\left(\vec{r}\right)\over\partial y}\right\rbrack^{2}
}\,
{\partial{\rm f}\left(\vec{r}\right)\over\partial y}
\end{array}\right.
$$
Equivalent to
$$
\delta\vec{r}
\approx
-\,{\nabla{\rm f}\left(\vec{r}\right)
 \over
 \left\vert\nabla{\rm f}\left(\vec{r}\right)\right\vert^{2}}
\,{\rm f}\left(\vec{r}\right)
$$
