Critical Points, hessian determinant null

[zxcvbs]

Hi im try to guess how to study the behaviour of functions, when the determinant on the second derivative test is null.
For example:

\(\displaystyle \[\begin{array}{l}
f(x,y) = {x^3} + {y^3} - 3x\\
\nabla f(x,y) = (3{x^2} - 3,3{y^2})\\
\left\{ \begin{array}{l}
3{x^2} - 3 = 0\\
3{y^2} = 0
\end{array} \right\}\\
Cp = ( - 1,0);(1,0)\\
Hf\left| {\begin{array}{*{20}{c}}
{6x}&0\\
0&{6y}
\end{array}} \right|\\
h{f_{(1,0)}}\left| {\begin{array}{*{20}{c}}
6&0\\
0&0
\end{array}} \right|\\
h{f_{( - 1,0)}}\left| {\begin{array}{*{20}{c}}
{ - 6}&0\\
0&0
\end{array}} \right|
\end{array}\]\)

I know i have to approximate the critical point by curves, to know its behaviour.
For example for (-1,0)

\(\displaystyle \[\begin{array}{l}
for\,\,( - 1,0)\\
\alpha (x) = P + t(1,0)\\
a(x) = ( - 1 + t,0)\\
f( - 1 + {t^2},0) = {( - 1 + t)^3} - 3( - 1 + t)\\
\alpha (x) = P + {t^2}(1,0)\\
a(x) = ( - 1 + {t^2},0)\\
f( - 1 + {t^2},0) = {( - 1 + {t^2})^3} - 3( - 1 + {t^2})\\
\alpha (y) = P + t(0,1)\\
a(y) = ( - 1,t)\\
f( - 1,t) = {( - 1)^3} + {t^3} - 3( - 1)\\
f( - 1,t) = {t^3} + 2
\end{array}\]\)

For y^3 i know is a cubic, and it doesnt have a max or a min, is an inflection point.
But for x= ( - 1 + t,0), i dont know how to study it.
I had seen other examples where looking just for y or x, gives you a saddle point (i.e. x^2-y^2).
But how can i study with functions when the hessian determinant is null.

Thanks.