Use the level curves of the function to determine if each partial derivative at the point P is positive, negative, or zero. 
It is asking me to find $f_{xx}$, $f_{yy}$, $f_x$, $f_y$, but I'm really unsure of how to determine the characteristics of the partial derivative.
Please give me some guidance, thanks.
 A: Let $P = (a, b)$. Define $g(u,v) = f(a + u, b + v)$.
Then, we may make the following approximations:


*

*$f_x (a, b) = g_x(0, 0) \approx g(1, 0) - g(0, 0)$

*$f_{xx} (a, b) = g_{xx}(0, 0) \approx g_x(1, 0) - g_x(0, 0)$


You can use method (1) to compute the $g_x$ terms:
$g_{xx}(0, 0)\approx (g(2,0) - g(1, 0)) - (g(1, 0) - g(0, 0))$
You can use any $dx$ you like, but I chose $dx = 1$ for simplicity.
A: There's one level of analysis you can use to get intuition without calculation. 
Geometrically, the gradient is perpendicular to the level curves and in the direction of greatest change, so this will be uniform in direction, up and to the left at a 45 degree angle to the x axis. 
The change in scalar $f(x,y)$ is $df=\nabla f \cdot d\vec{s}$ where $d\vec{s}$ is a line element representing the magnitude and direction of differential displacement.
We know $df$ is a maximum with $d\vec{s}$ is up and to the left. So $|df|=||\nabla f||$ 
$\partial f/\partial x=\nabla f \cdot\hat{i}$ where $\hat{i}$ is the unit vector in the $x$ direction. We don't exactly know the magnitude of the gradient, but we know the direction. specifically, we know the angle between the x axis and the gradient is 45 degrees. This means $\partial f/\partial x = ||\nabla f|| cos(45^\circ)$.By similar arguments one can find $\partial f/\partial y$. 
Note the level curves get closer and closer together as you go up and to the left. So our level curves are not just increasing as we go up and to the left. They are increasing at an increasing rate. 
So $\partial/\partial f = \hat{i}\cdot \nabla. $
To get $f_{xx}$, we want  $\hat{i}\cdot \nabla(\nabla f \cdot\hat{i})$
So $f_{xx}=(\cos{45^\circ})^2|| \nabla(||\nabla f||)||$
The same thing can be done with y. 
