<s>[INST] $f:(x,y)\mapsto \frac{x\sin(y)-y\sin(x)}{x^2+y^2}$ 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 ?
 [/INST] A partial answer to at least show that $f$ is differentiable at $(0,0)$.  Then it remains to show that its derivative is continuous.  I didn't check if that also follows from this bound directly.
Combine $\tan(x)\geq x$ and $\sin(x) \leq x$ for $x \in [0,\pi/2)$ to get
$$
\cos(x) \leq \frac{\sin(x)}{x} \leq 1
$$
for $x \in (-\pi/2,\pi,2)$.  Then for $x,y \in (-\pi/2,\pi/2)$
$$
-\frac{y^2}{2} \leq \cos(y)-1 \leq \frac{\sin(y)}{y} - \frac{\sin(x)}{x} \leq 1 - \cos(x) \leq \frac{x^2}{2}.
$$
Taking the absolute value:
$$
\left| \frac{\sin(y)}{y} - \frac{\sin(x)}{x} \right| \leq \frac{\max(x^2,y^2)}{2}\leq \frac{x^2+y^2}{2}.
$$
This results in the following estimate of $f$:
$$
\left|\frac{x\sin(y) - y\sin(x)}{x^2+y^2}\right| \leq \left| \frac{xy}{x^2+y^2}\right| \cdot \left| \frac{\sin(y)}{y} - \frac{\sin(x)}{x}\right| \leq \frac{x^2+y^2}{4}
$$
This is sufficiently sharp to conclude that $f$ is differentiable in $(0,0)$.
</s>