Second partial derivative: What happens if $f_{xy}^2$ is larger than $f_{xx} f_{yy}$ Considering $(a,b)$ is a critical point of a funciton $f(x,y)$ and
$D(x,y) = det(H(f(x,y))) =f_{xx}(x,y)f_{yy}(x,y) - (f_{xy}(x,y))^{2}$ is the determinant of the hessian matrix from that function.
If $D(a,b ) > 0$ then $f_{xx}(a,b)$ and $f_{yy}(a,b)$ both have to be either positive or negative, which leads to a local minimum or maximum.
If $D(a,b) < 0$ then $f_{xx}(a,b)$ and $f_{yy}(a,b)$ have opposite signs, which leads to a saddle point.
But what if $(f_{xy}(a,b))^{2} > f_{xx}(a,b)f_{yy}(a,b)$ then $D(x,y)$ would also be negative even if $f_{xx}(a,b)$ and $f_{yy}(a,b)$ have the same sign.
So what is the critical point in that case? Do i just look if $f_{xx}(a,b)f_{yy}(a,b)$ have the same sign or does the determinant matter?

That's also a saddle point and an interesting one. As Will Jagy suggested, the function will go up along both $x$ and $y$, but it's still a saddle and it will go downward along some tilted dimensions.
For example $f(x,y)=x^2+6xy+y^2$  its hessian matrix is
$$
H=\begin{bmatrix}
2 &6\\
6 &2\\
\end{bmatrix}
$$
Obviously $\det H<0$. As you see below, the function is an upward parabola both along $x$ and $y$, but along the diagonal it's a downward paraola. That's the geometric picture of a saddle point: some directions will lead upward, some downward.

