Find $\frac{\partial^2 f}{\partial x \partial y}(0, 0)$ given that $\lim_{(x, y) \to (0,0)} \frac{f(x, y) - \tan{(x)}\sin{(y)}}{x^2 + y^2} = 0$. $f \in C^2(\mathbb{R^2})$ satisfies $$\lim_{(x, y) \to (0, 0)} \frac{f(x, y) - \tan{(x)}\sin{(y)}}{x^2 + y^2} = 0.$$ Find $\frac{\partial^2 f}{\partial x \partial y}(0, 0).$
I've tried to deduce something from the limit and definition of partial derivatives but with no effect.
 A: As $f\in C^2({\mathbb R}^2)$ we know that there are constants $c_0$, $c_1$, $c_2$, $c_{11}$, $c_{12}$, $c_{22}$ such that
$$f(x,y)=c_0+c_1x+c_2y+c_{11}x^2+c_{12}xy+c_{22}y^2+o\bigl(x^2+y^2\bigr)\qquad\bigl((x,y)\to(0,0)\bigr)\ .$$
As
$$\tan x\>\sin y=xy+o(x^2+y^2)\qquad\bigl((x,y)\to(0,0)\bigr)$$
we see that
$$f(x,y)-\tan x \>\sin y =c_0+c_1x+c_2y+c_{11}x^2+(c_{12}-1)xy+c_{22}y^2+o\bigl(x^2+y^2\bigr)$$ when $(x,y)\to(0,0)$. With your limit assumption this implies $c_0=c_1=c_2=c_{11}=c_{22}=0$ and $c_{12}=1$, so that
$${\partial f^2\over\partial x\partial y}(0,0)=1\ .$$
A: You can expand to 
\begin{equation}
\frac{\partial^2 f(x,y)}{\partial x\partial y} = \frac{\partial\tan(x)}{\partial x} \frac{\partial\sin(y)}{\partial y}\\\approx \partial_x \partial_y \left(1- \frac{1}{2}y^2\right)\left(1+x^2\right)=-2xy
\end{equation}
All higher terms go to $0$, since they are $O(x^3,y^3)$. The initial condition fixes the order $O(x^0,y^0)$
Therefore you could say that to second leading order
$f(x,y) = -2xy$
So your second derivative equals to $-2$
