How do I differentiate (x^2 + 3x + 3)/(x+3)

In order to differentiate this fraction, you will need to use the quotient rule, which if you don't remember is:
y = u/v then dy/dx = (vdu/dx − u dv/dx)/ v^2)
First you need to identify your initial variables:
u = x^2 + 3x + 3
v = x+3
Then you need to differentiate these initial variables, which is done by multiplying the coefficient of each term by the number of the power and then subtracting one off of the power:
du/dx = 2x + 3dv/dx = 1
Remember that for number terms, the real term is 3x^0. This means that when you multiply by the power, entire term is turned to 0.
Now that you have your variables, all you have to do is substitute them into the formula:
dy/dx = (((x+3)(2x+3)) - ((x^2 + 3x + 3)*1))/(x + 3)^2
After you expand the brackets, this becomes
dy/dx = (2x^2 +6x + 3x + 9 - x^2 - 3x - 3)/(x+3)^2
dy/dx = (x^2 + 6x + 6)/(x+3)^2

This is the simplest form, therefore we have the answer!