Example 1 Given that $f(x)$ is an odd function defined on $[-6,6]$, and $f(x)$ is a linear function of $x$ on $[0,3]$, and a quadratic function of $x$ on $[3,6]$. When $3 \leqslant x \leqslant 6$, $f(x) \leqslant f(5)=3$, $f(6)=2$, find the analytical expression of the function $f(x)$.
Analysis Since $f(x)$ is an odd function on $[-6,6]$, we only need to find the expression of $f(x)$ on $[0,6]$ first. Then divide $[0,6]$ into two intervals $[0,3]$ and $[3,6]$, and find the analytical expressions of $f(x)$ respectively, the key is to find $f(3)$.
Solution Since $f(x)$ is a quadratic function on $[3,6]$, and $f(5)$ is its maximum value, we can set
$$
f(x)=a(x-5)^{2}+3,3 \leqslant x \leqslant 6,
$$

So
$$
2=f(6)=a(6-5)^{2}+3 .
$$

Solving the equation, we get $a=-1$.
Therefore, $f(x)=-(x-5)^{2}+3,3 \leqslant x \leqslant 6$.
From the above, we know that $f(3)=-1$. Since $f(x)$ is an odd function, $f(0)=0$, let
$$
\begin{array}{c}
f(x)=k x, 0 \leqslant x \leqslant 3, \\
-1=f(3)=3 k .
\end{array}
$$

Solving the equation, we get $k=-\frac{1}{3}$.
Therefore, $f(x)=-\frac{1}{3} x, 0 \leqslant x \leqslant 3$.
Since $f(x)=-f(-x)$, we have
When $-6 \leqslant x \leqslant-3$, $3 \leqslant-x \leqslant 6$, then
$$
f(x)=-f(-x)=-\left[-(-x-5)^{2}+3\right]=(x+5)^{2}-3 ;
$$

When $-3 \leqslant x \leqslant 0$, $0 \leqslant-x \leqslant 3$, then
$$
f(x)=-f(-x)=-\left(\frac{1}{3} x\right)=-\frac{1}{3} x .
$$

In summary, the analytical expression of $f(x)$ is
$$
f(x)=\left\{\begin{array}{ll}
(x+5)^{2}-3, & \text { when }-6 \leqslant x<-3 \text {, } \\
-\frac{1}{3} x, & \text { when }-3 \leqslant x<3 \text {, } \\
-(x-5)^{2}+3, & \text { when } 3 \leqslant x \leqslant 6 \text { . }
\end{array}\right.
$$

Note If the domain $D$ of an odd function includes the number 0, then it must be that $f(0)=0$, because $f(0)=-f(-0)$, thus $f(0)=0$.

From the definitions of odd and even functions, we know that the graph of an odd function is symmetric about the origin, and the graph of an even function is symmetric about the $y$-axis.