Let f(x)=x^3-3x. The number of distinct real solutions of f(f(x))=0 is

Since $f(t)=0$ has roots $t=0,\pm\sqrt3$, we need $f(x)\in\{0,\sqrt3,-\sqrt3\}$. The graph of $f$ has a local maximum $2$ at $x=-1$ and a local minimum $-2$ at $x=1$. For any constant $c$ with $|c|<2$, the horizontal line $y=c$ meets the cubic in exactly three points. Because $0$, $\sqrt3$ and $-\sqrt3$ all lie in $(-2,2)$ (as $\sqrt3\approx1.73$), each equation $f(x)=c$ gives three distinct real roots. Different values of $c$ give different $x$, so the total is $3\times3=9$.