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 $x=-y$

This is prossible?
$ 0\leq|xy\left(\frac{1+xy}{x^3+y^3}\right)^{1/3}|=|\left(\frac{x^3y^3+x^4y^4}{x^3+y^3}\right)^{1/3}|=|\left(\frac{x^3y^3}{x^3+y^3}+\frac{x^3y^3(xy)}{x^3+y^3}\right)^{1/3}|=|\left(\frac{x^3}{x^3+y^3}y^3+\frac{x^3}{x^3+y^3}y^3(xy)\right)^{1/3}|\leq|(y^3+y^3(xy))^{1/3}|\leq (y^2|y|+y^4|x|)^{1/3}\leq(|y|+|x|)^{1/3}$
so like the function is bounded by two function whose limits tends to $0$ then $f(x,y)$ tends to $0$¿?
