Question

$$3x^{2}y+3y^{3}=3$$

Solve for x

$x=\sqrt{-y^{2}+\frac{1}{y}}$
$x=-\sqrt{-y^{2}+\frac{1}{y}}\text{, }y>0\text{ and }y\leq 1$

Solve for y

$y=\frac{18^{\frac{2}{3}}\left(\sqrt[3]{\sqrt{3\left(4x^{6}+27\right)}+9}+\sqrt[3]{-\sqrt{3\left(4x^{6}+27\right)}+9}\right)}{18}$