f(x)=x2(x+3)3,
f'(x)=2x(x+3)3+3x2(x+3)2=0 at turning points:
x(x+3)2(2x+6+3x)=0,
x(x+3)2(5x+6)=0 when x=0, x=-3, x=-6/5.
(-∞,-3] f(x) is increasing up to inflexion point at x=-3;
[-3,-6/5] f(x) is increasing up to maximum (concave down) at x=-6/5;
[-6/5,0] f(x) is decreasing to minimum (concave up) at x=0;
[0,∞) f(x) increases continuously.