$x^{2}+6x+3y+16=y^{2}+3x+\frac{3}{x^{2}}\text{ and }y^{2}+3x+\frac{3}{x^{2}}=3$

$\left\{\begin{matrix}y=-\frac{\sqrt{4x^{4}+12x^{3}+73x^{2}-12}}{2x}+\frac{3}{2}\text{, }&4x^{4}+12x^{3}+73x^{2}-12\geq 0\text{ and }-12x-\frac{12}{x^{2}}\geq -12\text{ and }\frac{|\sqrt{4x^{4}+12x^{3}+73x^{2}-12}-3x|}{2|x|}=\frac{\sqrt{-3x^{3}+3x^{2}-3}}{|x|}\\y=\frac{\sqrt{4x^{4}+12x^{3}+73x^{2}-12}}{2x}+\frac{3}{2}\text{, }&4x^{4}+12x^{3}+73x^{2}-12\geq 0\text{ and }-12x-\frac{12}{x^{2}}\geq -12\text{ and }\frac{|\sqrt{4x^{4}+12x^{3}+73x^{2}-12}+3x|}{2|x|}=\frac{\sqrt{-3x^{3}+3x^{2}-3}}{|x|}\end{matrix}\right.$