Finding the extrema of $f(x,y) = 2x^{4} - 3x^{2}y + y^{2}$ $$f(x,y) = 2x^{4} - 3x^{2}y + y^{2}$$
I found the stationary points of this function using the equations -
$$\frac{\partial f}{\partial x} = 0 \qquad \frac{\partial f}{\partial y} = 0$$
I got $(0,0)$.
Now we calculate $R$, $S$, and $T$ at $(0,0)$ where 
$$R = \frac{\partial^{2} f}{\partial x^{2} } \qquad
S = \frac{\partial^{2} f}{\partial x\partial y} \qquad
T = \frac{\partial^{2} f}{\partial y^{2} }$$
then, 
$$RT - S^{2} = 0$$
This does't lead to any result. How to find nature of this stationary point?
I know what local minima, local maxima, and saddle points are, but how to find the nature of $(0,0)$?

$$f(x,y)=2x^4-3x^2y+y^2=\left( y-2 {{x}^{2}}\right) \, \left( y-{{x}^{2}}\right)$$
$$f\left( \varepsilon ,\frac{3 {{\epsilon }^{2}}}{2}\right) =-\frac{{{\varepsilon }^{4}}}{4}<0$$
$$f\left( \varepsilon ,\frac{{{\varepsilon }^{2}}}{2}\right) =\frac{3 {{\varepsilon }^{4}}}{4}>0$$
Then $(0,0)$ is saddle point.
