Finding the minimum or maximum of a bivariate function when $f_{xx}\times f_{yy}-f_{xy}^2=0$. I see that when $f_{xx}\times f_{yy}-f_{xy}^2<0$ then it is a saddle point.
Also when $f_{xx}\times f_{yy}-f_{xy}^2>0$ then it is a minima or maxima.
What is exactly happening when $f_{xx}\times f_{yy}-f_{xy}^2=0$?

An indefinite Hessian is a sufficient but not necessary condition for a stationary point to be a saddle point. For example the function given by by $x^4-y^4$ has a (unique) stationary point $(x,y)=(0,0)$ at which the determinant of the Hessian is zero.
Consider also the functions given by $(x+y)^2$, $-(x+y)^2$, $(x+y)^3$. In all cases the determinant of the Hessian is zero for all real numbers $x$ and $y$. In the first case the minimum is zero and there are many minimizers ($(t,-t)$ for any $t$) which are all stationary points. In the second case the maximum is zero and there are many maximizers ($(t,t)$ for any $t$) which are all stationary points. In the third case there are no maximizers or minimizers, but there are lots of stationary points ($(t,-t)$ for any $t$).  These are all saddle points.

In case you are not aware of what a Hessian is: the Hessian matrix arrays the second order derivatives:
$\begin{pmatrix}f_{xx}(x,y)& f_{xy}(x,y)\\ f_{yx}(x,y) & f_{yy}(x,y)\end{pmatrix}$
The determinant of Hessian is negative ($f_{xx}f_{yy}-f_{xy}^2<0$) if and only if the Hessian is indefinite.
