If f(x)= x^3+ax^2+bx, then
f '(x)= 3x^2+2ax+b and
f "(x)= 6x+2a.
f(x) has an inflection point at x= -1, which means that f "(x)=0 when x=-1. So f "(x)= 6x+2a=0 when x=-1
6(-1)+2a=0
-6+2a=0
2a=6
a=3.
f(x) also has a minimum at x=3, which means that f '(x)=0 when x=3. So,
f '(x)= 3x^2+2ax+b=0 when x=3
3(3)^2+2a(3)+b=0
We already have the value for a, so we can substitute it as well.
27+6(3)+b=0
45+b=0
b=-45.
It isn't really necessary to consider f''(x). See my post above.
All you have to know is that the given x coordinates are the coordinates of stationary points, and hence are solutions to the equation f'(x) = 0. You have two of these points, and two unknown variables, so it is simply to formulate two equations which can be solved simultaneously to find the values of a and b.