How I can evaluate $\lim_{(x,y) \rightarrow (0,0)} xy(\frac{1+xy}{x^3+y^3})^{1/3}$ I don't have idea how I can evaluate this double limit 
$$\lim_{(x,y) \rightarrow (0,0)} xy \left(\frac{1+xy}{x^3+y^3} \right) ^{1/3}$$ could you help me please!



I try prove that $f$ is continuous: $f(x,y)=xy \left(\frac{1+xy}{x^3+y^3} \right) ^{1/3}$ if $x\not=-y$ and $f(x,y)=0$ if $x=-y$
Hint: replace $1+xy$ by $1$ in the numerator (why?) and reduce to
$$
\lim_{(x,y) \to (0,0)} \left( \frac{x^3 y^3}{x^3+y^3} \right)^{1/3}.
$$
Put $\alpha =x^3$, $\beta=y^3$ and consider the constraint $\alpha + \beta = \alpha^3$ with $\alpha \to 0$. Then
$$
\frac{\alpha \beta}{\alpha+\beta} = \frac{\alpha^4-\alpha^2}{\alpha^3}
$$
which becomes unbounded as $\alpha \to 0$.
Try Polar coordinates maybe $x=r \cos \phi$ and $y=r \sin \phi$ and then check the limit $r \to 0$.