differential equation

solve, 3y' (y'')^2 + ( (y')^2 - 1 ) ( (y')^3 - y' - y''' ) = 0 , where y'=dy/dx ,
y''=d^2 y /dx^2 , y'''=d^3 y/dx^3 , and y is a function of x (I have found a particular solution, siny=ksinx, where k is a constant