x³ + y³ = 3axy---(1) at (3a/2, 3a/2)

\(\therefore\) 3x² + 3y² \(\frac{dy}{dx}\) = 3a(x\(\frac{dy}{dx}\) + y)----(2)

(by differentiating given equation w.r.to x)

⇒ (3y² - 3ax)\(\frac{dy}{dx}\) = 3ay - 3x²

⇒ \(\frac{dy}{dx}\) = \(\frac{3ay-3x^2}{3y^2-3ax}\) 

⇒ \((\frac{dy}{dx})\)_{(x = 3a/2, y = 3a/2)} = \(\cfrac{\frac{9a^2}2-3\times\frac{9a^2}4}{3\times\frac{9a^2}4-\frac{9a^2}2}\) 

 = \(\frac{18a^2-27a^2}{27a^2-18a^2}=\frac{-9a^2}{9a^2}=-1\)

Differentiating equation(2) w.r.to x, we get

6x + 6y\((\frac{dy}{dx})^2\) + 3y²\(\frac{d^2y}{dx^2}\) = 3a(x\(\frac{d^2y}{dx^2}\) + \(\frac{dy}{dx}+\frac{dy}{dx}\))

⇒ (3y² - 3ax)\(\frac{d^2y}{dx^2}\) = -6y\((\frac{dy}{dx})^2\) - 6x - 6a\(\frac{dy}{dx}\)

⇒ \(\frac{d^2y}{dx^2}\)(at x = 3a/2, y = 3a/2) = \(\cfrac{-6\times\frac{3a}2\times(-1)^2-6\times\frac{3a}2-6a\times-1}{3\times\frac{9a^2}4-3a\times\frac{3a}2}\) 

 = \(\cfrac{-9a-9a+6a}{\frac{27a^2-18a^2}4}\) = \(\frac{-12a\times4}{9a^2}=\frac{-16}{3a}\)

\(\therefore\) Radius of curvature of given curve at (3a/2, 3a/2)

 = \(\left|\cfrac{(1+(\frac{dy}{dx})^2)^{3/2}}{\frac{d^2y}{dx^2}}\right|\) = \(\left|\cfrac{(1+(-1)^2)^{3/2}}{\frac{-16}{3a}}\right|\)

 = \(|\frac{-2\sqrt2\times3a}{16}|\) = \(|\frac{-3\sqrt2a}8|\) = \(\frac{3\sqrt2a}{8}\)(\(\because\) Radius is never negative)