Relationship between second order derivatives and cross derivative of smooth surfaces Probably a silly question, but I wonder if $z=f(x,y)$ is a smooth surface, and the values of its two second order derivatives $\frac{\partial^2f}{\partial x^2}$ and $\frac{\partial^2f}{\partial y^2}$ are known to be both positive throughout a region, say $0<x<1$, $0<y<1$, dose this place a restriction on the possible values of the cross derivative, i. e. $\frac{\partial^2f}{\partial x\partial y}$ ? 
The naive geometrical idea behind this is: if I cut a surface horizontally (parallel to the $x$-axis) or vertically (parallel to the $y$-axis) and always find an exponential (or other) curve, does this makes a random cut obliquely more likely to be a exponential-like (or other-like) curve ?
 A: I think your intuition is a bit off. The quantity $f_{xx}$  tells you about the shapes of cross-section curves lying in planes $y = \text{constant}$, as you said. Similarly, $f_{yy}$ tells you about cross-sections in a perpendicular direction. It seems that this ought to tell you something about oblique sections, I agree. But the flaw in your reasoning is to associate $f_{xy}$ with these oblique cross-sections. In fact, $f_{xy}$ tells you how the slopes of cross-section curves change as you move the sectioning plane. In other words, it tells you about variation of cross-sections, not about the cross-sections themselves.
There is a related theorem that might be of interest: Euler's formula tells you how the curvatures of planar sections vary as you rotate the sectioning plane around a surface normal. In fact, the curvature in any direction can be calculated very simply from the curvatures in two specific directions (called the principal directions) that happen to be perpendicular to each other. Not exactly what you had in mind, but in the same spirit, I think.
A: Your original question is not as far-fetched as you think. To elaborate on @Jesse Madnick's response, to understand the behavior (i.e., curvature) of the various cross-sections, you need all the Hessian matrix. That is, if the tangent plane to your surface is horizontal, the second-order Taylor polynomial is 
\begin{align*}
f(x,y) &= f(0,0) + Q(x,y)\,, \quad\text{where}\\
Q(x,y) &= \tfrac12\left(f_{xx}(0,0)x^2 + 2f_{xy}(0,0)xy + f_{yy}(0,0)y^2\right)\,.
\end{align*}
That is, the curvature of the slice in direction $\vec v$ is obtained by putting $\vec v$ into the quadratic form given by the Hessian matrix.
