The sign of the second order derivative sign Suppose that we know the sign of $\frac{\partial f(x,y)}{\partial x}$ and $ \frac{\partial f(x,y)}{\partial y}$ on a specific domain. Can we say that the sign of $ \frac{\partial^2 f(x,y)}{\partial x \partial y}$ is basically the product of the signs of the first order derivatives?
Example: $f(x,y)=\frac{1}{x y}$ :   $\frac{\partial f(x,y)}{\partial x}\leq 0$ and $ \frac{\partial f(x,y)}{\partial y}\leq 0$ and  $ \frac{\partial^2 f(x,y)}{\partial x \partial y}\geq 0$ for all $x\geq 0$ and $y\geq 0$.
If this is correct, do you have a source? My search engine attempts didn't give me something :-(
Thanks a lot
 A: A simple counterxample is provided by a function $f(x,y)=xy$ on a domain where $x<0<y$. Then $\frac{\partial f(x,y)}{\partial x}=y>0$,
$\frac{\partial f(x,y)}{\partial y}=x<0$, whereas $\frac{\partial^2  f(x,y)}{\partial x\partial y}=1>0$.
A: Your example works for another reason: $f(x,y)=g(x)h(y)$ where $g'(x)<0$ and $h'(y)<0,$ so the mixed partial has sign $(-1)(-1)=1.$ In case of functions of the form $f(x,y)=g(x)h(y),$ you're on the right track because
$$\frac{\partial^2 f(x,y)}{\partial x \partial y} = g'(x)h'(y),$$
while $$\frac{\partial f}{\partial x} \frac{\partial f}{\partial y}= g(x)h(y)g'(x)h'(y),$$
which has the same sign as $g'(x)h'(y)$ if $g(x)$ and $h(y)$ have the same sign on your domain. But notice that domains make the conjecture still false in general (see Alex's answer!). So unfortunately the conjecture is nowhere near to being true, even if the domain is such that $f(x,y)$ is positive. For example, consider $f(x,y)=\sin(xy)$ on $0<x,y\le \sqrt{\pi}$ (so that $0< xy\le \pi$). You have that 
$$\frac{\partial f}{\partial x}=y\cos(xy), \frac{\partial f}{\partial x}=x\cos(xy), \frac{\partial^2 f}{\partial x\partial y}=-xy\sin(xy).$$
The sign of $f_{xy}=-xy\sin(xy)$ is the same as the sign of $-\sin(xy)$ in our domain: it's always negative. However, $f_x=-y\cos(xy)$ and $f_y=-x\cos(xy)$ have the same sign as $-\cos(xy),$ which can be positive or negative. In both cases $f_x\cdot f_y $ is always positive, having the same sign as $\cos^2(xy)$ on our domain.
