The correct options are
A Points of equilibrium are x=1, x=2 and x=3
B x=2 is point of unstable equilibrium
C x=1 is point of stable equilibrium
D x=3 is point of stable equilibrium
F=−dUdx=−x3+6x2−11x+6
For equilibria, F = 0 ⇒x3−6x2+11x−6=0
x3−3x2−3x2+9x+2x−6=0
x2(x−3)−3x(x−3)+2(x−3)=0
(x3−3x+2)(x−3)=0⇒(x−1)(x−2)(x−3)=0
Now dF=−(3x2−12x+11)dx
at x=2, dF=+dx→ unstable
at x=1, dF=−2dx→ stable
at x=3, dF=−2dx→ stable