Find the Area!

The set $S'$ is just a square of side length 2 that's centered at the origin of the $xy-$ plane and has vertices at $(-1, \pm 1); (1, \pm 1)$ .

Taking the derivative of the governing nonlinear equation for the set $S$ produces:

$\frac{d}{dx} (2xy - 3x^{2}y = C) \Rightarrow (2y + 2xy') - (6xy + 3x^{2}y') = 0 \Rightarrow y' = \frac{2y(3x-1)}{x(2-3x)};$

which we require $x \ne 0, \frac{1}{3}, \frac{2}{3}$ AND $y \ne 0$ in order to satisfy $y' > 0.$ Upon examination of $S \cap S'$ , its total area is just the set of all points $(x,y)$ contained within:

$S_1 = (x,y): x \in [-1,0); y \in (0,1] \Rightarrow$ AREA = $(1)(1) = 1$

$S_2 = (x,y): x \in (0, \frac{1}{3}); y \in [-1,0) \Rightarrow$ AREA = $(\frac{1}{3})(1) = \frac{1}{3}$

$S_3 = (x,y): x \in (\frac{1}{3}, \frac{2}{3}); y \in (0,1] \Rightarrow$ AREA = $(\frac{1}{3})(1) = \frac{1}{3}$

$S_4 = (x,y): x \in (\frac{2}{3}, 1]; y \in [-1,0) \Rightarrow$ AREA = $(\frac{1}{3})(1) = \frac{1}{3}$

or $S_1 + S_2 + S_3 + S_4 = 1 + 3(\frac{1}{3}) = \boxed{2}.$