Question

Find all the critical values

Answer 1

              y - 3
g(y) = ------------------
            y^2 - 3y + 9

To find the c. v., we must differentiate this function g(y) and set the derivative equal to zero:
     
             (y^2 - 3y + 9)(1) - (y - 3)(2y - 3)
g '(y) = --------------------------------------------
                    (y^2 - 3y + 9)^2

Note carefully:  The denom. has no real roots, so division by zero is not going to be an issue here.  

Simplifying the denominator of the derivative, 

(y^2 - 3y + 9)(1) - (y - 3)(2y - 3)  =>  y^2 - 3y + 9 - [2y^2 - 3y - 6y + 9], or
                                                         -y^2 + 6y

Setting this result = to 0 produces the equation y(-y + 6) = 0, so
y = 0 and y = 6.  These are your critical values.  You may or may not have max or min at one or the other.