The function is defined such that at $(x,y)=(0,0)$ we have that $f(x,y)=0$. And so I was trying to prove that $\lim_{(x,y)\rightarrow(0,0)}\frac{x^2\sin(xy)}{\sqrt{x^2+y^2}}=0$ and am not sure whether my bound is correct, would appreciate some feedback on the proof.
Proof:
Let $\epsilon>0$ be arbitrary. Let $\delta=\epsilon^{\frac{1}{3}}$ such that when $\vec{x}=(x,y)\in{}D$ and $0<||\vec{x}-\vec{0}||<\delta$ we have, \begin{equation} \left|\frac{x^2\sin(xy)}{\sqrt{x^2+y^2}}-0\right|=\frac{|x^2\sin(xy)|}{\sqrt{x^2+y^2}}=\frac{|x^2xy\cdot\sin(xy)|}{\sqrt{x^2+y^2}\cdot{}xy} \end{equation}
Whereas $\frac{\sin(xy)}{xy}\rightarrow1$ as $(x,y)\rightarrow(0,0)$ and hence it leaves us with,
\begin{equation} \frac{|x^3y|}{\sqrt{x^2+y^2}}=\frac{|x|^3|y|}{||\vec{x}||}\leq\frac{||\vec{x}||^3||\vec{x}||}{||\vec{x}||}=||\vec{x}||^3<\delta^3=\epsilon. \end{equation}
How does this look?