Find area bounded by $(\frac{x}{3}+\frac{y}{6})^3-9xy=0$ $C = \{(x,y) \mid (\frac{x}{3}+\frac{y}{6})^3-9xy=0\}$.
I want to find the area bounded by $C$ in first quadrant.
Can you tell me how to solve exercises like this?
I have no idea what integral I need to solve, because my curve is implicitly defined.

Let's use (slightly modified) polar coordinates. Set $x = 3r\cos\theta$ and $y = 6r\sin\theta$. 
The Jacobian of this transformation is $\left|\begin{matrix}\tfrac{\partial x}{\partial r} & \tfrac{\partial y}{\partial r} \\ \tfrac{\partial x}{\partial \theta} & \tfrac{\partial y}{\partial \theta}\end{matrix}\right| = \left|\begin{matrix}3\cos\theta & 6\sin\theta \\ -3r\sin\theta & 6r\cos\theta\end{matrix}\right| = 18r$. 
The curve $C$ in $(r,\theta)$ coordinates is given by: 
$\left(\dfrac{3r\cos\theta}{3}+\dfrac{6r\sin\theta}{6}\right)^3-9(3r\cos\theta)(6r\sin\theta) = 0$
$r^3(\cos\theta+\sin\theta)^3 - 162r^2\cos\theta\sin\theta = 0$
$r = \dfrac{162\cos\theta\sin\theta}{(\cos\theta+\sin\theta)^3} =: R(\theta)$
Thus, the area of the region bounded by $C$ is: 
$\displaystyle\iint\limits_{C}\,dx\,dy$ $=\displaystyle\int_{0}^{\pi/2}\int_{0}^{R(\theta)}18r\,dr\,d\theta$ $=\displaystyle\int_{0}^{\pi/2}9R(\theta)^2\,d\theta$ $=\displaystyle\int_{0}^{\pi/2}\dfrac{9 \cdot 162^2\cos^2\theta\sin^2\theta}{(\cos\theta+\sin\theta)^6}\,d\theta$ $=\displaystyle9 \cdot 162^2\int_{0}^{\pi/2}\dfrac{\tan^2\theta\sec^2\theta}{(1+\tan\theta)^6}\,d\theta$ $= 9 \cdot 162^2\displaystyle\int_{0}^{\infty}\dfrac{u^2}{(1+u)^6}\,du$ $= 9 \cdot 162^2\displaystyle\int_{1}^{\infty}\dfrac{(v-1)^2}{v^6}\,dv$ $= 9 \cdot 162^2\displaystyle\int_{1}^{\infty}\left(\dfrac{1}{v^4}-\dfrac{2}{v^5}+\dfrac{1}{v^6}\right)\,dv$ $= 9 \cdot 162^2 \left[-\dfrac{1}{3u^3}+\dfrac{1}{2u^4}-\dfrac{2}{5u^5}\right]_{1}^{\infty}$ $= 7873.2$
