Inconclusive Second derivative test ,Now how shall i proceed While doing maxima and minima questions i have encountered upon question in which i cannot show nature of points 
P1 : Given $f(x,y) = 2x^4 - 3x^{2}y + y^{2}$
Doubtful case is as origin .
P2: $f(x,y)$ = $y^{2} + x^{4} +x^{2}y $
Doubtful case is at origin
Thanks for help ..
 A: P1: The function is $0$ on the curves $y = x^2$ and $y = 2 x^2$, is negative in between them and positive otherwise. So $(0,0)$ is a saddle point since there are positive and negative values of $f$ arbitrarily close to $(0,0)$. The fact that it is a saddle does not follow simply from the discriminant being $0$, for example the origin is a minimum for $g(x,y,)=x^4+y^4$.
A: $P_1: \dfrac{\partial f}{\partial x} = 4x^3-6xy = 0, \dfrac{\partial f}{\partial y} = 2y-3x^2 = 0 \Rightarrow 4x^3-3x(3x^2)= 0 \Rightarrow -5x^3=0 \Rightarrow x = 0 \Rightarrow y = 0$.
Next, $\dfrac{\partial^2 f}{\partial x^2}|_{(0,0)} = 12x^2 -6y|_{(0,0)} = 0$
$\dfrac{\partial^2 f}{\partial x\partial y}|_{(0,0)} = -6x|_{(0,0)} = 0$
$\dfrac{\partial^2 f}{\partial y^2}|_{(0,0)} = 2 \Rightarrow \left|\begin{matrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{matrix}\right| = \left|\begin{matrix} 0 & 0 \\ 0 & 2\end{matrix}\right| = 0$. 
Thus for $P_1$, we have a saddle point at the origin. I hope you can continue solving $P_2$....
