$f:(x,y)\mapsto \frac{x\sin(y)-y\sin(x)}{x^2+y^2}$ is a $C^1$-function I would like to show that the function:
$$f:(x,y)\mapsto \frac{x\sin(y)-y\sin(x)}{x^2+y^2}$$ 
is a $C^1$-function.
$$ \frac{\partial f}{\partial x}(x,y)=\frac{\sin(y)-y\cos(x)}{x^2+y^2}+\frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2}$$
$$ \frac{\partial f}{\partial y}(x,y)=-\frac{\partial f}{\partial y}(y,x)=... $$
So I just have to show that:
$$ \frac{\partial f}{\partial x}(x,y)\rightarrow_{(0,0)}0$$
When $y\geq0$ :
$$ -\frac{y^3}{6(x^2+y^2)}+\frac{x^2y}{x^2+y^2}-\frac{x^4y}{4!(x^2+y^2)} \leq \frac{\sin(y)-y\cos(x)}{x^2+y^2} \leq \frac{yx^2}{2(x^2+y^2)}$$
When $y<0$ :
$$ -\frac{y^3}{6(x^2+y^2)}+\frac{y^5}{5!(x^2+y^2)}+\frac{x^2y}{2(x^2+y^2)} \leq \frac{\sin(y)-y\cos(x)}{x^2+y^2} \leq \frac{yx^2}{2(x^2+y^2)}-\frac{yx^4}{4!(x^2+y^2)}$$
So $$ \frac{\sin(y)-y\cos(x)}{x^2+y^2}\rightarrow_{(0,0)}0 $$
How can I directly find an upper bound of $$ \left| \frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2} \right|$$ that tends to 0 ?

Because of the Taylor expansion $\sin(x) = x - {x^3 /6} + ...$, if $x$ and $y$ are small enough you can write $\sin(x) = x + E(x)$ and $\sin(y) = y + E(y)$, where $|E(x)| < |x|^3$ and $|E(y)| < |y|^3$. So you have
$${2x(y\sin(x) - x\sin(y)) \over (x^2 + y^2)^2} = {2x(yx + yE(x) - xy - xE(y)) \over (x^2 + y^2)^2}$$
$$= {2x(yE(x) - xE(y)) \over (x^2 + y^2)^2}$$
Taking absolute values and bounding, this is at most
$$= 2|x|{|yE(x)| + |xE(y)| \over (x^2 + y^2)^2}$$
Inserting $|E(x)| < |x|^3$ and $|E(y)| < |y|^3$ this is bounded by
$$= 2|x|{|x^3y| + |xy^3| \over (x^2 + y^2)^2}$$
$$= 2|x|{|xy|(x^2 + y^2) \over (x^2 + y^2)^2}$$
$$= 2|x| {|xy| \over x^2 + y^2}$$
I think you can take it from there...
