Let $y=tx,$ after calculation, I get
$$x=\frac{3at}{1+t^3}, y=\frac{3at^2}{1+t^3}.$$
Use the Green Formula, the area is equal to
$$\frac{1}{2}\oint_{\Gamma}xdy-ydx.$$
My question is: how to determine the range of $t$? If I don’t use a drawing software, I wouldn’t even know what this curve looks like. How to determine the range of $t$ for the closed curve on the graph？
Thanks for advance！
First of all, let us show that for any angle $0 < \theta < \frac{\pi}{2}$, there is a unique $r(\theta) > 0$ such that $x^3 + y^3 = 3axy$ (I assume $a > 0$) with $x = r(\theta)\cos(\theta)$ and $y = r(\theta)\sin(\theta)$. Indeed, such an $r$ should verify the following equation, $$ r^3(\cos^3(\theta) + \sin^3(\theta)) = 3ar^2\cos(\theta)\sin(\theta). $$ It easily gives $r(\theta) = 3a\frac{\cos(\theta)\sin(\theta)}{\cos^3(\theta) + \sin^3(\theta)} > 0$. We deduce that the part of the curve in the upper right part of the plane is parametrized by the curve, $$ \gamma : \theta \mapsto (r(\theta)\cos(\theta),r(\theta)\sin(\theta)), $$ with $0 \leqslant \theta \leqslant \frac{\pi}{2}$. You dedcue that, $$ x(\theta) = r(\theta)\cos(\theta) = 3a\frac{\cos^2(\theta)\sin(\theta)}{\cos^3(\theta) + \sin^3(\theta)} = 3a\frac{\tan(\theta)}{1 + \tan^3(\theta)}, $$ and similarly $y = 3a\frac{\tan^2(\theta)}{1 + \tan^3(\theta)}$. To obtain the wanted parametrization, set $t = \tan(\theta)$. Since $\theta$ varies from $0$ to $\frac{\pi}{2}$, $t$ varies from $0$ to $+\infty$. And, \begin{align*} \frac{1}{2}(xdy - ydx) & = \frac{1}{2}(x(t)y'(t) - y(t)x'(t))dt\\ & = \frac{9a^2}{2}\left(\frac{t}{1 + t^3}\frac{2t(1 + t^3) - t^2 \cdot 3t^2}{(1 + t^3)^2} - \frac{t^2}{1 + t^3}\frac{1 + t^3 - t \cdot 3t^2}{(1 + t^3)^2}\right)dt\\ & = \frac{9a^2}{2}\frac{-t^5 + 2t^2 + 2t^5 - t^2}{(1 + t^3)^3}dt\\ & = \frac{9a^2}{2}\frac{t^5 + t^2}{(1 + t^3)^3}dt\\ & = \frac{9a^2}{2}\frac{t^2}{(1 + t^3)^2}dt. \end{align*} Notice that you could have used $xdy$ or $-ydx$ instead (by Stoke's theorem). Finally, \begin{align*} \mathrm{Area}(\Omega) & = \frac{9a^2}{2}\int_0^\infty \frac{t^2}{(1 + t^3)^2}dt\\ & = \frac{9a^2}{2}\left[-\frac{1}{3(t^3 + 1)}\right]_0^\infty\\ & = \frac{3a^2}{2}. \end{align*}