Find $x^2+y^2$ if $x^2-\frac{2}{x}=3y^2$ and $y^2-\frac{11}{y}=3x^2$ 
Find $x^2+y^2$ if $x^2-\frac{2}{x}=3y^2$ and $y^2-\frac{11}{y}=3x^2$.

My try:
$$\frac{y^2-\frac{11}{y}}{3}-\frac{2}{x}=3y^2$$
then?

Assume that $x,y\in\mathbb{R}$ and $a,b\in\mathbb{R}$ satisfy $$x^2-\frac{a}{x}=3y^2\text{ and }y^2-\frac{b}{y}=3x^2\,.$$
Then, taking $z:=x+\text{i}y$, we have
$$z^3=\left(x^3-3xy^2\right)-\text{i}\left(y^3-3x^2y\right)=a-b\text{i}\,.$$
Thus,
$$x^2+y^2=|z|^2=\sqrt[3]{\left|z^3\right|^2}=\sqrt[3]{|a-b\text{i}|^2}=\sqrt[3]{a^2+b^2}\,.$$
Caveat: This solution only works if $x$ and $y$ are real numbers.  If they can be complex numbers, dxiv's first solution is better.  In fact, his solution shows that $x^2+y^2$ can be any of the three cubic roots of $a^2+b^2$.
