Given
2x³ + (x³ - 3) sin(2πy) - 3y = 0
we first notice that when x = ³√3, we get
2 (³√3)³ + ((³√3)³ - 3) sin(2πy) - 3y = 0
2•3 + (3 - 3) sin(2πy) - 3y = 0
6 - 3y = 0
3y = 6
y = 2
Differentiating both sides with respect to x gives
6x² + 3x³ sin(2πy) + 2π (x³ - 3) cos(2πy) y' - 3y' = 0
Then when x = ³√3, we find
6(³√3)² + 3(³√3)³ sin(2π•2) + 2π ((³√3)³ - 3) cos(2π•2) y' - 3y' = 0
6•³√9 + 3•3 sin(4π) + 2π (3- 3) cos(4π) y' - 3y' = 0
6•³√9 + 0 + 0 - 3y' = 0
3y' = 6•³√9
y' = 2•³√9
(that is, 2 times the cube root of 9)
+
·sin(2πy)−3y = 0.