Example 4 Given real numbers $a, b$ satisfy
$$
a^{2}+a b+b^{2}=1 \text {, and } t=a b-a^{2}-b^{2} \text {. }
$$

Then the range of values for $t$ is
From the given conditions, we have
$$
\begin{array}{l}
a^{2}+b^{2}=\frac{1-t}{2}, \\
4 a^{2} b^{2}=(1+t)^{2} .
\end{array}
$$

From equation (1), we can set
$$
a^{2}=\frac{1-t}{4}+p, b^{2}=\frac{1-t}{4}-p \text {. }
$$

Substituting into equation (2) and simplifying, we get
$$
(t+3)(3 t+1)=-16 p^{2} \text {. }
$$

Therefore, $(t+3)(3 t+1) \leqslant 0$.
Solving this, we get $-3 \leqslant t \leqslant-\frac{1}{3}$.