The function $f(x,y)=2y^3-6y^2+3x^2y$ has 2 stationary points, $(0,0)$ and $(0,2)$. Using the function's Hessian I managed to prove that $(0,2)$ is a strict local minima, but the Hessian of $f$ at $(0,0)$ is $6\left({\begin{array}{cc} 0&0\\0&-2 \end{array} } \right)$ which is negative semi-definite, and by looking at the surface plot of the function + checking with WolframAlpha I'm pretty sure it's a saddle point and not a local maxima (although WolframAlpha did not classify it as a saddle point, but also not as a local maxima).
$f(0,0)=0$, and I did manage to prove that for $\epsilon>0$ the value $f(-\epsilon,-\epsilon)$ is negative. I'm trying to find a combination of expressions using $\epsilon$ that when plugged into the function gives out a value which is always positive, so as to prove that each neighborhood of the point contains both positive and negative (smaller and larger) values.
I need help proving/disproving $(0,0)$ is a saddle point/local maxima. Thanks!
Edit: fixed the function
As you said, $$ f(x,x) = 5x^3 - 6 x^2 $$ is negative for $x$ small enough. On the other hand, for $y > 0$, $$ f(\sqrt{2y},y) = 2 y^3 $$ is positive.