Is the function $xy-(\sin{y})(\sin{x} +2y)$ non-positive near $(0,0)$? As in the title is the function $f(x,y)=xy-(\sin{y})(\sin{x} +2y)$ non-positive near $(0,0)$? 

The wolfram claims so - see the plot. On the other hand one can argue as follows. Choose arbitrary small $x&gt;0$ since $x&gt; \sin x$ one can find $y&gt;0$ small enough to have $x&gt; \sin x +2y$ since also $y&gt; \sin y$ one easel gets $xy \ge(\sin{y})(\sin{x} +2y)$ and wolfram confirms that also for some concrete numbers - see this. What am I missing? Thanks in advance for Your help.
If you expand as Taylor series around $x=0$
$$A=x y-\sin (y) (\sin (x)+2 y)=-2 (y \sin (y))+x (y-\sin (y))+\frac{1}{6} x^3 \sin (y)+O\left(x^4\right)$$ Peform another expansion of the previous around $y=0$ to get for $y &lt;&lt; x$ $$A\approx \frac 16 x^3 y$$ which matches the number in the second run.

Using, as in the second run, $x=10^{-3}$ and computing with illimited precision, what we can notice is that $A &gt;0$ as soon as $y &lt; 10^{-11}$.
Considering limit from all sides
You get,
$$\lim_{(x,y) \to (0,0)} \lfloor xy-\sin y (\sin x +2y)\rfloor = 0$$

Which indicates the fact that,
$$\lim_{(x,y)\to (0,0) } f(x,y) \to 0^{+}$$

Thus,
The function is positive near the given the point.