(i) y'''=(y')^3
Let z=y', then z"=y''' and z"=z^3.
z^-3d(dz/dx)/dx=1. Using S to denote integral: S(z^-3d(dz/dx))=S(dx).
Integrating: -(1/2)z^-2(dz/dx)=x+a where a is constant of integration.
Integrating again: -(1/2)S(z^-2dz)=S((x+a)dx)=x^2/2+ax+b where b is a constant of integration.
So, (1/2)z^-1=x^2/2+ax+b; multiply through by 2z: 1=z(x^2+2ax+2b).
But z=dy/dx, so dy/dx=1/(x^2+2ax+2b) and, integrating again:
y=S(dx/(x^2+2ax+2b))=S(dx/((x+a)^2+2b-a^2)).
Let A^2=2b-a^2 and X=x+a so dX=dx and y=S(dX/(X^2+A^2)).
Now if X=Atan(t), dX=Asec^2(t)dt and
y=S(Asec^2(t)dt/(A^2(tan^2(t)+1))=S(Asec^2(t)dt/(A^2sec^2(t))=(1/A)S(dt)=t/A=tan^-1(X/A)/A+c
Substituting for X and A: y=tan^-1((x+a)/sqrt(2b-a^2))/sqrt(2b-a^2)+c, where c is integration constant.
If 2b<a^2 the square root has a negative argument, leading to a complex solution.
(ii) y"=(y')^2
Let z=y' then z'=y" and z'=z^2, so z^-2z'=1.
Integrating: S(z^-2dz)=S(dx)=x+a.
-z^-1=x+a so multiplying through by -z: 1=-z(x+a).
z=dy/dx=-1/(x+a) and y=-S(1/(x+a))=-ln(x+a)+ln(b)=ln(b/(x+a)); y=ln(b/(x+a)) or e^y=b/(x+a), where b is constant.