\(\displaystyle s(t)= 3t^2- t^3= t^2(3- t)\). \(\displaystyle t^2\) is non-negative for all t and 0 only for t= 0, 3- t= -(t- 3) is positive for t< 3, 0 at t= 3, and negative or t>3. So for t< 0, s(t) is the product of two positive numbers so is positive. s(t)= 0. s(t) is the product of two positive number so is positive again for 0< t< 3. s(3)= 0. s(t) is the product of positive and a negative so is negative for t> 3.
So, the graph, as we go from negative to positive, comes down from the top left corner of the graph ("from infinity) to (0, 0) where it is tangent to the s-axis, goes back up to some maximum value, then back down to (3, 0) and then down to the lower right corner of the graph (to negative infinity.
Do you understand that the velocity function and acceleration function are the first and second derivatives of \(\displaystyle s(t)= 3t^2- t^3\)? Do you know what those derivatives are?
It is impossible to tell what they are asking for because you have not told us what the question is! Apparently you are given the function s(t)= 3t^2- t^3 and are asked to graph it and its derivative, s'(t)= 6t- 3t^2. Is that right? If so then the graphs in your second post are correct.
For III you calculate the derivative of \(\displaystyle f(x)= 3x^2\) at x= 2 using the limit of the difference quotient. Is that what the problem asked you to do? Yes, that derivative is 6(2)= 12.