Let's determine the minimum and maximum values of the expression $9-x^{2}-2 \sqrt{9-x^{2}}$, where $x$ is a real number between -3 and 3.
Given the interval $-3 \leq x \leq 3$, the expression is defined. Let's introduce the notation $a=\sqrt{9-x^{2}}$. Then,

$$
f(x)=9-x^{2}-2 \sqrt{9-x^{2}}=a^{2}-2 a
$$

is the quadratic function $g(a)=a^{2}-2 a=(a-1)^{2}-1$, which has its minimum at $a=1$. From the equation $\sqrt{9-x^{2}}=1$, we get $x_{1,2}= \pm 2 \sqrt{2}$, and $f\left(x_{1,2}\right)=-1$. Since $2 \sqrt{2}$ and $-2 \sqrt{2}$ both fall within the given interval, the expression has a minimum at both points, and the minimum value is -1.

The function $g(a)=(a-1)^{2}-1$ is strictly decreasing on the interval $[0,1]$ and strictly increasing on the interval $[1,3]$. Therefore, it can only achieve its maximum at $a=0$ or $a=3$. If $a=0$, then from $\sqrt{9-x^{2}}=0$ we get $x_{3,4}= \pm 3, f( \pm 3)=0$. If $a=3$, then from $\sqrt{9-x^{2}}=3$ we get $x_{5}=0$, and $f(0)=9-2 \sqrt{9}=3$. This is the maximum value of the expression in the given interval.

Remark. Many solved the problem using derivatives, but some forgot that where the derivative is zero, the function may have an extremum. To confirm that there is indeed an extremum, it must be checked separately.