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 $x>0$ since $x> \sin x$ one can find $y>0$ small enough to have $x> \sin x +2y$ since also $y> \sin y$ 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.
On
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.
On
Claim. The function $f$ takes both signs in the neighborhood of $(0,0)$.
Proof. Consider the pullback of $f$ along the curve $y=x^4$, i.e., the function $$\eqalign{g(x)&:=f(x,x^4)=x^5-\sin(x^4)(\sin x + 2 x^4)\cr &=x^5-\left(x^4-{x^{12}\over 6}+?x^{20}\right)\left(x-{x^3\over6}+2x^4+?x^5\right)\cr &={x^7\over6}+?x^8\qquad(x\to0)\ ,\cr}$$ where the $?$ acts as standin for various functions that are analytic near $x=0$. It follows that $${\rm sgn}\bigl(g(x)\bigr)={\rm sgn}(x)\qquad (x\to0)\ .$$
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 $y << x$ $$A\approx \frac 16 x^3 y$$ 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 $A >0$ as soon as $y < 10^{-11}$.