using impliciate deferation find dy/dx for y^2+X^3=3XY
y^2 + x^3 = 3xy
differentiating both sides wrt x,
d(y^2)/dx + d(x^3)/dx = 3.d(xy)/dx
Taking the three differentials individually,
d(y^2)/dx = d(y^2)/dy * dy/dx (using chain rule)
d(y^2)/dx = 2y * y’ (where y’ = dy/dx)
d(x^3)/dx = 3x^2 (normal differentiation rule)
3.d(xy)/dx = 3(1.y + x.y’) (using product rule)
Summing these three results appropriately,
2y.y’ + 3x^2 = 3(y + x.y’)
2y.y’ + 3x^2 = 3y + 3x.y’
2y.y’ – 3x.y’ = 3y – 3x^2
(2y – 3x).y’ = 3(y – x^2)
y’ = dy/dx = 3(y – x^2)/(2y – 3x)