Number of line of curvature meets at one point I am now consider the surface given by $$f=(x,y,x^3-3xy^2)$$
And I have been asked to prove that there are three line of curvature meet at origan point.
I can compute that
\begin{align*}
  f_x&=(1,0,3x^2-3y^2)\\
  f_y&=(0,1,-6xy)\\
  f_x\times f_y&=(-3(x^2-y^2),6xy,1)\\
  |f_x\times f_y|&=\sqrt{9(x^2-y^2)^2+36x^2y^2+1}=\sqrt{9(x^2+y^2)^2+1}\\
  n&=\frac{f_x\times f_y}{|f_x\times f_y|}=\frac{(-3(x^2-y^2),6xy,1)}{\sqrt{9(x^2+y^2)^2+1}}\\
  f_{xx}&=(0,0,6x)\\
  f_{xy}&=(0,0,-6y)\\
  f_{yy}&=(0,0,-6x)\\
  E&=<f_x,f_x>=1+9(x^2-y^2)^2\\
  F&=<f_x,f_y>=-18(x^2-y^2)xy\\
  G&=<f_y,f_y>=1+36x^2y^2\\
  L&=<n,f_{xx}>=\frac{6x}{\sqrt{9(x^2+y^2)^2+1}}\\
  M&=<n,f_{xy}>=\frac{-6y}{\sqrt{9(x^2+y^2)^2+1}}\\
  N&=<n,f_{yy}>=\frac{-6x}{\sqrt{9(x^2+y^2)^2+1}}.\\
 \end{align*}
When $x=y=0$, we have that $$(g)=\begin{pmatrix}
  E & F\\
  F & G
 \end{pmatrix}=\begin{pmatrix}
  1 & 0\\
  0 & 1
 \end{pmatrix}\mbox{ and }(b)=\begin{pmatrix}
  L & M\\
  M & N
 \end{pmatrix}=\begin{pmatrix}
  0 & 0\\
  0 & 0
 \end{pmatrix}.$$
Hence the shape operator $$(s)=(g^{-1})(b)=\begin{pmatrix}
  0 & 0\\
  0 & 0
 \end{pmatrix}.$$
So the principal curvatures at point $x=y=0$ is that $k_1=k_2=0$.
I do not know how to continuous and I think I can not even calculate the principal direction at the origan and how can I know how the number of lines of curvature.
But I have observed that this surface is mirror symmetrical to $(x,z)-$plane. Does it have any relations?
I have also calculated that a line of curvature $f\circ \gamma$ with $\gamma=\begin{pmatrix}
\gamma_1\\
\gamma_2
\end{pmatrix}$ should fullfill the following equation $$\begin{vmatrix}
\gamma_2^{\prime 2} & -\gamma_1^\prime \gamma_2^\prime & \gamma_2^{\prime 2}\\
E & F & G\\
L & M & N \end{vmatrix}=0$$
but how can I know a line of curvature cross the origan point?

Principal directions at $(x,0)$ (for $x\ne 0$) are $(1,0)$ and $(0,1)$.
The computations are very difficult unless you happen to recognize that $x^3-3xy$ is the real part of $(x+iy)^3$. Thus, this surface has rotational symmetry about the origin, with rotations of $\pm 2\pi/3$. If one does the computation with a polar coordinate parametrization, these things become more evident. So the $x$-axis and its $\pm 2\pi/3$ rotated images give the three lines of curvature coming into the origin.
P.S. Without any hints, I think this problem is rather unfair.
