Number of line of curvature meets at one point

Question

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?

Answer 1

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.