calculus - Is $f(x,y)=frac{sinsqrt[3]{x^3+y^3}}{sqrt[5]{x^5+y^5}}$ uniformly continuous or not

Find out if function $$f(x,y)=\frac{\sin\sqrt[3]{x^3+y^3}}{\sqrt[5]{x^5+y^5}}$$ is uniformly continous or not in area $D=\{0I found out that we have no $\lim\limits_{x,y\to 0}f(x,y)$, because $$\lim\limits_{x,y\to 0}\frac{\sin\sqrt[3]{x^3+y^3}}{\sqrt[5]{x^5+y^5}}=\lim\limits_{\rho\to0}\frac{\sqrt[3]{\rho^3\sin^3\alpha+\rho^3\cos^3\alpha}}{\sqrt[5]{\rho^5\sin^5\alpha+\rho^5\cos^5\alpha}}=\lim\limits_{\rho\to0}\frac{\sqrt[3]{\sin^3\alpha+\cos^3\alpha}}{\sqrt[5]{\sin^5\alpha+\cos^5\alpha}}$$
hence we can't use The Uniform Continuity Theorem as we can't determ $f(0,0)$. Function doesn't have bounded partial derivatives, so I think it's not uniformly continous, but I don't know how to show that