consider the function
$$ f(x,y) = \dfrac{x^{1/3}}{(x+ay^3)^{1/3}} $$ where $a>0$ is a constant and $x,y\geq 0$. It is easy to see that, outside of the origin, $0\leq f\leq 1$. $f$ itself is not continuous (consider the restriction on the $x,y$ axis and then consider limit towards the origin), but $xyf(x,y)$ should be and should also be differentiable, but with a differential that is not bounded on $[0,+\infty)\times [0,+\infty)$.
Now, I have the following expression
$$ \Delta = x_1y_1f(x_1,y_1) - x_2y_2f(x_2,y_2). $$
My question is: can we get a bound on $\Delta$ of the form
$$ \Delta\leq A|x_1-x_2|+B|y_1-y_2|\quad \mbox{or}\quad \Delta\leq C|x_1-x_2||y_1-y_2| $$
with $A,B,C$ constants? If not (as I think), can we get at least a bound of the form
$$ \Delta\leq A|x_1-x_2|^\alpha+B|y_1-y_2|^\beta\quad \mbox{or}\quad \Delta\leq C|x_1-x_2|^\alpha|y_1-y_2|^\beta $$ or a combination of those, with $\alpha\leq 4/3$? Note:, if $\beta>0$, then B must be as small as possible.
Edit: It would be enough to get a bound on $\Delta\cdot(y_1-y_2)$.
It seems the following.
Mean Value Theorem suggests that $\Delta=\nabla(xyf)\cdot (x_2-x_1,y_2-y_1)$, where a point $(x,y)$ belongs to the segment connecting points $(x_1, x_2)$ and $(y_1, y_2)$. I obtained with Mathcad $$\nabla(xyf)=\left(\frac{\partial(xyf)}{\partial x},\frac{\partial(xyf)}{\partial y}\right)= \frac{x^{1/3}}{(x+ay^3)^{4/3}}\left(xy+\frac 43 ay^4, x^2\right).$$
Now restrict $x_2=x_1+1$ and $y_2=y_1+1$. If one of your bounds hold, then both coordinated of the vector $\nabla(xyf)$ should be bounded, which, as you already suggested, is not true. Indeed, if $y_1$ is constant and $x_1$ tends to infinity then the second coordinate of $\nabla(xyf)$ tends to infinity too.