Multivariable function that satisfy some conditions [closed]

Let $f(x,y)$ be a differentiable function that is not identically zero and for all $x,y>1,$

(1) $f(x,x)=0,$

(2) $f_x(x,y) f_y(x,y) \geq 0$ for all $x,y$ sufficiently large,

(3) $\lim \limits_{x \to \infty, y \to \infty} f(x,y)=0.$

Does such a function exist?

$$f(x,y) = \begin{cases} -\mathrm{e}^{4 \left(\frac{1}{x^2-4} + \frac{1}{y^2 -4} \right)}(x-y)^2,& (x,y) \in [1,2]\times[1,2] \\ 0 ,& \text{otherwise} \end{cases}$$

The particular $f$ here is a bump function, stretched by a factor of $2$ (so naturally sitting on $[-2,2] \times [-2,2]$) times $(x-y)(y-x)$ to get the necessary negative sign (so we can subsequently be increasing) and to obtain condition (1).

The exponential is never zero and the "$(x-y)^2$" ensures condition (1). Condition (2) holds trivially for $x \geq 2$ or for $y \geq 2$. Condition (3) holds trivially. $f(3/2,5/4) = \frac{-1}{16 \mathrm{e}^{61/16}} \neq 0$. $f$ is certainly differentiable outside of $(1,2] \times (1,2]$ and inside $(1,2) \times (1,2)$, so the only thing left to show is that $f$ is differentiable along $\{2\}\times (1,2]$ and along $(1,2] \times \{2\}$. The convenient property of this bump function is that its derivatives of all orders (including order zero) are zero on those two line segments, which matches exactly the derivatives of the zero function on the other side.

One might object that this is the zero function with nearly irrelevant non-zero-ness near $(1,1)$. This is true, but also unavoidable. Note that condition (2) requires that the function is (non-strictly) monotonically increasing in the (positive) axis-parallel directions once the coordinates are sufficiently large. Now pick an $x$ that is sufficiently large. At $(x,x)$, (1) requires that $f(x,x) = 0$, (2) that $f$ is (non-strictly) increasing in the $y$-direction, and $(3)$ that it never leaves any arbitrarily small neighborhood of $0$ (since monotonicity ensures it can never return). So, for all large enough $x$, $f$ is zero on the ray $\{x\} \times [x,\infty)$. Similarly, for all large enough $y$, $f$ is zero on the ray $[y,\infty) \times \{y\}$. Consequently, $f$ is required to be identically zero on $[a,\infty) \times [b,\infty)$ for some constants $a,b \in (1,\infty)$. Equivalently, the conditions require that $f$ can only be nonzero on $\{(x,y) \in (1,\infty)^2 \mid x < a \text{ or } y < b\}$ with the same conditions on $a$ and $b$.