2. Given that the function $f(x)$ is an odd function on $\mathbf{R}$, and when $x \geqslant 0$, $f(x)=x^{2}$, then the solution set of the inequality $f(f(x))+$ $f(x-1)<0$ is $\qquad$
$$
f(f(x))<-f(x-1)=f(1-x) \Rightarrow f(x)<1-x \Rightarrow\left\{\begin{array}{l}
x^{2}<1-x, x \geqslant 0, \\
-x^{2}<1-x, x<0
\end{array}\right.
$$
$\Rightarrow x<\frac{\sqrt{5}-1}{2}$, so the solution set of the original inequality is $\left(-\infty, \frac{\sqrt{5}-1}{2}\right)$.