We have y3−3y+x = 0 Differentiating both sides, we get 3y2y′−3y′+1 = 0 (1) Substituting x = 22 , we get y = - 102. ⇒ y′(−102) = rac−12 Differentiating Eq. (1), we get 3y2y′′+6y(y′)2−3y′′ = 0 Substituting x = 22 , we get y = −102 and y' = rac−12 Therefore, y′′(−102) = rac4273imes32