Show that $f(x,y)=\dfrac{\sin(x^4-y^4)}{x^2+y^2}$ is steady on $\Bbb{R}$ My task is to show that 
$$f(x,y) =
\begin{cases}
\dfrac{\sin(x^4-y^4)}{x^2+y^2},  & \text{if $(x,y) \ne (0,0)$} \\[2ex]
0, & \text{else}
\end{cases}$$
is steady on $\Bbb{R}$
As a composition of steady functions, it is enough to show that $f(x,y)$ converges to $0$ for $(x,y) \rightarrow (0,0)$
I've started with the following:
$$\tag{$\forall(x,y)\ne(0,0)$}x = \cos(\phi)\cdot r$$$$y = \sin(\phi)\cdot r$$
$$f(x,y) = f(\phi, r) = \dfrac{\sin(r^4(cos^4(\phi)-\sin^4(\phi)))}{r^2(\cos^2(\phi)+\sin^2(\phi))}$$
$$=\dfrac{\sin(r^4(\cos^4(\phi)-\sin^4(\phi)))}{r^2}$$
With $$\lim_{(x,y)\rightarrow(0,0)}f(x,y)= \lim_{r\rightarrow 0}f(\phi, r)$$
It can be seen that the term converges to $0$ for $r\rightarrow 0$ but I wanted to show this mathematically:
$$\lim_{r\rightarrow 0}f(\phi, r) = \lim_{r\rightarrow 0}\dfrac{d f(\phi, r)}{\color{red}{d?}} = \color{red}{?}$$
My question is:
w.r.t which variable would I need to take the derivative in order to do this mathematically correct.
I am very happy for any kind of help or advice.
Greetings, Finn

$|f(x,y)| \le \frac{|x^4-y^4|}{x^2+y^2} \le \frac{x^4+y^4}{x^2+y^2}  \le x^2+y^2$.
Can you proceed ?
