Consider the curve defined by x3+y3=3xyx^3 + y^3 = 3xyx3+y3=3xy. Find the value of d2ydx2\frac{d^2y}{dx^2}dx2d2y at the point (32,32)(\frac{3}{2}, \frac{3}{2})(23,23).
−3227-\frac{32}{27}−2732
−827-\frac{8}{27}−278
−23-\frac{2}{3}−32
000