Example 5 Given real numbers $a, b$ satisfy $a^{2}+a b+b^{2}=1$, and $t=a b-a^{2}-b^{2}$. Then the range of $t$ is $\qquad$ $t_{\max }=$ $\qquad$ ,$t_{\text {min }}=$ $\qquad$ .
(2001, TI Cup National Junior High School Mathematics Competition)
Let $a=x+y, b=x-y$.
Substitute into the given equation:
$$
(x+y)^{2}+(x+y)(x-y)+(x-y)^{2}=1 \text {. }
$$

Simplify to get $y^{2}=1-3 x^{2}$.
Since $y^{2} \geqslant 0$, we have $0 \leqslant x^{2} \leqslant \frac{1}{3}$.
Thus, $t=a b-a^{2}-b^{2}$
$$
\begin{array}{l}
=(x+y)(x-y)-(x+y)^{2}-(x-y)^{2} \\
=-\left(x^{2}+3 y^{2}\right)=-x^{2}-3\left(1-3 x^{2}\right) \\
=8 x^{2}-3 .
\end{array}
$$
$$
\begin{array}{l}
\text { Clearly, } 0 \leqslant 8 x^{2} \leqslant \frac{8}{3} \\
\Rightarrow-3 \leqslant 8 x^{2}-3 \leqslant-\frac{1}{3} \\
\Rightarrow-3 \leqslant t \leqslant-\frac{1}{3} \Rightarrow t_{\max }=-\frac{1}{3}, t_{\min }=-3 .
\end{array}
$$