The problem is to find the supremum of the set of integrals of the form $$I=\int_C(y^3-y)dx-3x^3dy$$ where $C$ ranges over all positively oriented simple closed plane curves.
The idea I have is to apply Green's theorem to conclude that $I=\int\int_D (-9x^2-3y^2+1)dA$ where $D$ is the region bounded by $C$. I believe this integral has maximum value if $\int\int_D -(3x^2+y^2)dA$ does. But I don't know what to do next.
$\int\int_D (-9x^2-3y^2+1)dA$
The integral will be maximized if $-9x^2-3y^2+1 \ge 0$ for all $x,y\in D$
$9x^2 + 3y^2 \le 1$ is an ellipse
$x = \frac {r}3 \cos t\\y = \frac {r}{\sqrt 3} \sin t\\dy\ dx = \frac {r}{3\sqrt 3} \ dr\ dt$
$\int_0^{2\pi}\int_0^1 -\frac {r^3}{3\sqrt3}+\frac {r}{3\sqrt 3}\ dr\ dt$