$$x^{2/3}+y^{2/3}=a^{2/3}\longleftrightarrow\;x=a\cos^3t\;,\;\;y=a\sin^3t\;,\;\;0\le t\le 2\pi$$
So
$$\frac12\int\limits_0^{2\pi}\left[(a\cos^3t\cdot3a\sin^2t\cos t)-(a\sin^3t\cdot(-3a\cos^2t\sin t)\right]dt =$$
$$=\frac{3a^2}2\int\limits_0^{2\pi}(\cos^4t\sin^2t+\cos^2t\sin^4t)dt=\frac{3a^2}2\int\limits_0^{2\pi}\cos^2t\sin^2t\,dt=$$
$$=\left.\frac{3a^2}{64}\left(4x-\sin 4x\right)\right|_0^{2\pi}=\frac{3a^2}{64}8\pi=\frac38a^2\pi$$
The statement is true when $\gamma$ is a positively oriented simple closed curve bounding some Jordan domain $\Omega$.
Let $\mathcal{I}$ be the integral at hand. Identify Euclidean plane $\mathbb{R}^2$ with the complex plane $\mathbb{C}$ and introduce complex coordinates $x,y$:
$$
\begin{cases}
\vec{x} = (x_1,x_2) & \leftrightarrow & x = x_1 + x_2 i\\
\vec{y} = (y_1,y_2) &\leftrightarrow & y = y_1 + y_2 i
\end{cases}$$
Let $z = y - x$. For any $\vec{\rho} = (\rho_1,\rho_2) \leftrightarrow \rho = \rho_1 + i\rho_2 \in \mathbb{C}$, we will use the notation $\Omega + \rho$ and $\gamma + \rho$ to denote the image of $\Omega$ and $\gamma$ under translation $\vec{\rho}$.
In terms of complex coordinates, we have
$d\vec{x} \cdot d\vec{y} = \frac12 \left(dx d\bar{y} + dy d\bar{x}\right)$. This leads to
$$\begin{align}
\mathcal{I}
&= -\frac{1}{8\pi} \int_{\gamma \times \gamma} \log(z\bar{z}) (dx d\bar{y} + dyd\bar{x})
= -\frac{1}{4\pi}\Re \left[\int_{\gamma\times\gamma} \log(z\bar{z}) dx d\bar{y}\right]\\
&= -\frac{1}{4\pi}\Re\left[\int_{x \in \gamma} \left(\int_{z \in \gamma - x}\log(z\bar{z}) d\bar{z}\right) dx\right]
\end{align}\tag{*1}
$$
Apply Stoke's theorem (the version for complex coordinates) to the inner integral, we obtain
$$\begin{align}
\mathcal{I}
&= -\frac{1}{4\pi}\Re\left[\int_{x\in\gamma} \left(\int_{z \in \Omega - x}
\left(\frac{dz}{z} + \frac{d\bar{z}}{\bar{z}}\right)\wedge d\bar{z}\right) dx\right]\\
&= \frac{1}{4\pi}\Re\left[\int_{x\in\gamma} \left(
\int_{z\in \Omega - x}\frac{d\bar{z} \wedge dz}{z}
\right) dx\right]\\
&= \frac{1}{4\pi}
\Re\left[\int_{x\in\gamma} \left(
\int_{y\in \Omega}\frac{d\bar{y} \wedge dy}{y-x}
\right) dx\right]\\
&= \frac{1}{4\pi}
\Re\left[
\int_{y\in\Omega}\left(\int_{x\in\gamma}\frac{dx}{y-x}\right) d\bar{y} \wedge dy
\right]
\end{align}
$$
By Cauchy's integral formula, we have $$\int_{x\in\gamma}\frac{dx}{y-x} = -2\pi i\quad\text{ for } y \in \Omega$$ As a result,
$$\begin{align}\mathcal{I}
&= \frac{1}{4\pi} \Re\left[ \int_{y \in \Omega} (-2\pi i)(2i dy_1 \wedge dy_2 )\right]
= \Re\left[ \int_{y \in \Omega} dy_1 \wedge dy_2 \right]\\
&= \Re\left[\verb/Area/(\Omega)\right] = \verb/Area/(\Omega)
\end{align}
$$
Update - Generalization to non-simple closed curves.
For non-simple closed curve $\gamma$ and $y \not\in \gamma$, let $W(\gamma;y)$ be the winding number of $\gamma$ around $y$. We have following generalization of Cauchy integral formula:
$$\int_\gamma \frac{dx}{y-x} = -2\pi i W(\gamma;y)\tag{*2}$$
When $\gamma$ satisfies following conditions:
- $\gamma$ can be decomposed into finitely many curve segments, the segments either completely coincides or their interior (as a curve segment) are disjoint from each other.
- $\gamma$ divides $\mathcal{C}\setminus \gamma$ into finitely many connected components, the boundaries of these components are finite combinations of curve segments from step $1$.
We can break any contour integral over $\gamma$ to integral combinations of contour integrals over boundaries of connected components in step $2$.
We can apply Stoke's theorem to the individual components and recombine the results.
Apply this procedure to inner contour integral in $(*1)$ and with help of $(*2)$, we obtain:
$$
\mathcal{I}
= \frac{1}{4\pi}
\Re\left[\int_{x\in\gamma} \left(
\int_{y\in \mathbb{C}\setminus\gamma}\frac{W(\gamma;y)}{y-x} d\bar{y} \wedge dy
\right) dx\right]
= \int_{y\in \mathbb{C}\setminus\gamma} W(\gamma;y)^2 dy_1 \wedge dy_2$$
Recall winding number is constant over each connected component. If $\Omega_1, \ldots, \Omega_m$ are the connected components of $\mathbb{C}\setminus\gamma$ with non-zero winding numbers and $W_i$ is the winding number for $\Omega_i$,
we can rewrite last expression as
$$\mathcal{I} = \sum_{i=1}^m W_i^2 \verb/Area/(\Omega_i)$$
The integral $\mathcal{I}$ is simply a weighted sum of the areas of connected components and the weight equals to the square of corresponding winding number.
In the special case where all $|W_i| \le 1$, above formula simplifies to
$$\mathcal{I} = \sum_{i=1}^m \verb/Area/(\Omega_i)$$
$\mathcal{I}$ becomes the unsigned area of those connected components whose winding number is non-zero.
Best Answer
I assume that $z$ is $\gamma$ in the definition of $\operatorname{diameter}(+\gamma)$. Without loss of generality, we can assume that $\gamma(a)=(0,0)$ (otherwise, we can translate $\gamma$ until the point $\gamma(a)$ hits the origin).
I would take $\vec{F}=(F_1,F_2)$ with $F_1=-y$ and $F_2=x$, so that $$\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}=2.$$ By Green's theorem, we have $$2\operatorname{Area}(\omega)=\iint_\omega \left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)\ dx\ dy=\oint_{+\gamma}\vec{F}\cdot \vec{dr}.$$ Hence, $$2\operatorname{Area}(\omega)\leq \oint_{+\gamma}\left\vert\vec{F}\cdot \vec{dr}\right\vert\leq \oint_{+\gamma}\left\Vert\vec{F}\right\Vert\ dr.$$ Because $\left\Vert\vec{F}\right\Vert\leq \operatorname{diameter}(+\gamma)$, we conclude that $$2\operatorname{Area}(\omega)\leq \oint_{+\gamma}\operatorname{diameter}(+\gamma)\ dr=\operatorname{diameter}(+\gamma)\oint_{+\gamma}dr=\operatorname{diameter}(+\gamma)\operatorname{length}(+\gamma).$$
This inequality is a quite weak. If we define $\operatorname{radius}(+\gamma)$ to be $$\inf\Big\{r>0:\exists p\in\mathbb{R}^2,\ \forall u\in[a,b],\ \big\vert\gamma(u)-p\big\vert< r\Big\}\,,$$ then we have $$2\operatorname{Area}(\omega)\leq \operatorname{radius}(+\gamma)\operatorname{length}(+\gamma).$$ The equality holds iff $\gamma$ traces a circle (once). You can use Jung's theorem to show that $$2\operatorname{Area}(\omega)\leq \frac{1}{\sqrt{3}}\operatorname{diameter}(+\gamma)\operatorname{length}(+\gamma),$$ which is stronger than the required result, but is still weak. So, it is an interesting question to find the infimum $\lambda_{\min}$ of all $\lambda>0$ such that $$\operatorname{Area}(\omega)\leq \lambda\operatorname{diameter}(+\gamma)\operatorname{length}(+\gamma).$$ We already know that $\lambda_{\min} \leq \frac1{2\sqrt{3}}$.