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 $x=-y$
 A: 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 $\alpha \to 0$.
A: 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 $0$¿?
A: In any neighborhood of $(0,0)$ there is a point of the form $(a,-a)$ with $0<a<1.$ Think of what happens as we approach $(a,-a)$ from the right. We have
$$f(a+h,-a) =   [\dots ]\cdot \left ( \frac{1}{(a+h)^3 +(-a)^3}\right)^{1/3},$$
where the expression in brackets has a nonzero limit as $h\to 0^+.$ But the other factor looks like $1/0^+$ as $h\to 0^+.$ It follows that $f$ is unbounded in any neighborhood of $(0,0),$ and so it's not even close: the limit in question doesn't exist, bigtime.
A: Try Polar coordinates maybe $x=r \cos \phi$ and $y=r \sin \phi$ and then check the limit $r \to 0$.
