$y=-\frac{\left(\sqrt{2x}+6\right)^{2}}{8}+\frac{3}{4}$
$x\geq 0$

$x=\left(-\sqrt{3-4y}+3\sqrt{2}\right)^{2}$
$y=-\frac{15}{4}\text{ or }arg(-\sqrt{3-4y}+3\sqrt{2})\geq \pi $

$y=-\frac{\left(\sqrt{2x}+6\right)^{2}}{8}+\frac{3}{4}$
$arg(\sqrt{2x}+6)<\pi $

$x=\left(-\sqrt{3-4y}+3\sqrt{2}\right)^{2}$
$-\left(-\sqrt{3-4y}+3\sqrt{2}\right)\geq 0\text{ and }y\leq \frac{3}{4}$