saddle point geometry of $f(x,y)$ when partial derivatives are of same sign. The sign of Hessian matrix and sign of any partial derivative of function $f(x,y)$ gives the informations about maxima, minima, saddle points of function and sometimes perplexed informations when it is equal to zero. 
Now when I start to figure out the geometrical interpretation of these condition I can understand why the maxima?,why the minima?, and even why the saddle point? When $H(x,y)\leq0 $ and both the derivative are of opposite sign. But when the condition is $H(x,y)\leq 0$ and both the derivatives are of same sign then i can not understand why the sadddle point? and how it looks? please help in explaining this.
 A: First of all we observe that the function $f$ needs to be of class $C^2$. Now the Hessian matrix $H(f)$ is defined as
$$
H(f)=
\begin{pmatrix}
\partial_{xx}f&\partial_{xy}f\\
\partial_{yx}f&\partial_{yy}f
\end{pmatrix}.
$$
Since $f\in C^2$ by the Schwarz theorem $\partial_{xy}f=\partial_{yx}f$; in other words the matrix $H(f)$ is symmetric.
The determinant of $H(f)$ is equal to $d(x,y)=\partial_{xx}f\partial_{yy}f- (\partial_{xy}f)^2$, and if $p=(x_0,y_0)$ is a critical point, then we you have the following cases:
1- $d(p)>0$ and $p$ is a minimum if $\partial_{xx}f(p)>0$, or a maximum if $\partial_{xx}f(p)<0$
2- $d(p)<0$ then $p$ is a saddle point;
3- $d(p)=0$ and so you don't have enough information to say what kind of critical point $p$ is.
Now I give you geometric meaning of a critical point. If $f:\Omega\longrightarrow \mathbb{R}$ is a function of class $C^2$ as above and $\Omega\subseteq \mathbb{R}^2$ an open domain; then the plot $\Gamma(f)$ is a Riemannian surface. In particular the determinant of the hessian matrix coincide with the determinant of the matrix associate to second fundamental form $(EG-F^2)$.
In differential geometry, when you study riemannian surface (not a Riemann surface!!!) the determinant of the matrix associate to the second fundamental form gives you a classification of the point from the geometrical point of view.
A point $p$ in said to be:
1- elliptic if: $d(p)>0$ and the eingenvalues of the matrix are both positive;
2- hyperbolic if: $d(p)>0$ and the eigenvalues of the matrix are both negative;
3- saddle point if: $d(p)<0$, so the enginvalues have opposite signs,
4- flat point if: $d(p)=0$ so one of the eigenvalues is $0$.
As you can see this is essentially the same classification of the critical point from the analytical point of view. 
Infact if the Hessian matrix is positive define then its eigenvalues are both of them positive, or negative. In the first case $\partial_{xx}f(p)>0$ so, the point is a minimun. Suppose that $p=0$ for simplicity, then if you restrict $f$ to the $x$-axis, in other words we fix $y=0$; the the plot of $f(x,0)$ is convex. The same holds if you restrict $f$ to $y$-axis.
In the second case the point is a maximum and restricting the function over $x$ or $y$ axis, the plot is concave.
If the Hessian determinant is negative, then one of its eigenvalues is positive, say it $\lambda_1$ and restricting the function along the direction gives by the eigenvector relative to $\lambda_1$ the plot is convex; viceversa restricting the function along the direction relative to negative eigenvalue $\lambda_2$ the plot is concave.
Finally if $d(p)=0$ this mean that the geometry around $p$ is flat, in other words a neighborhood of $p$ is locally isometric to a plane, then you are not able to understand in $p$ is a local maximum or minimum or saddle point.
