Find maximum or minimum values of the functions

$y=9-(x-3{)}^2$

$y=9-(x-3{)}^2=9-x^2-9+6x$

$y=6x-x^2$

$\frac{dy}{dx}=6-2x$

For minimum or maximum value of y. We will substitute $\frac{dy}{dx}=0$

6 - 2x=0  or x=3

To check whether value of y is maximum or minimum at x=3 we will have to check whether $\frac{d^{2}y}{d{x}^2}$ is positive or negative.

$\frac{d^{2}y}{d{x}^2}=-2$

or $\frac{d^{2}y}{d{x}^2}$ is negative at x=3. Hence, value of y is maximum. This maximum value of y is

$y_{\max }=9-(3-3{)}^2=9$