The correct option is
A (2cos2u−1)sin2uGiven :
u=tan−1(x3+y3x−y)
tanu=x3+y3x−y
(x−y)tanu=x3+y3→(1)
Differentiate w.r.t ′x′ partially
+tanu+(x−y)sec2u∂u∂x=3x2→(2)
Differntiate w.r.t ′y′ partially
−tanu−(x−y)sec2u∂u∂x=3x2→(3)
Further x(2) + y(3)
(x−y)tan(u)+sec2u(x−y){x∂u∂x+y∂u∂y}=3(x3+y3)
(x−y)[tan(u)+sec2u{x∂u∂x+y∂u∂y}]=3(x3+y3)
tan(u)+sec2u{x∂u∂x+y∂u∂y}=3(x3+y3)(x−y)
tan(u)+sec2u{x∂u∂x+y∂u∂y}=3tanu
⇒x∂u∂x+y∂u∂y=sin(2u)
Differentiating w.r.t x and y gives
∂u∂x+x∂2u∂x2+x∂2u∂x∂y=2cos(2u)∂u∂x→(4)
x∂2u∂x∂y+y∂2u∂y2+∂u∂y=2cos(2u)∂u∂y
Further x(4) + y(5) gives
x2∂2u∂x2+2xy∂2u∂x∂y+y2∂2u∂y2={2cos(2u)−1}{x∂u∂x+y∂u∂y}
={2cos(2u)−1}sin(2u)
Hence option A is correct answer