We see the size of the set $f^{-1}\left(f^{-1}\left(f^{-1}\left(f^{-1}(3)\right)\right)\right)$. Note that $f(x)=(x-1)^{2}-1=3$ has two solutions: $x=3$ and $x=-1$, and that the fixed points $f(x)=x$ are $x=3$ and $x=0$. Therefore, the number of real solutions is equal to the number of distinct real numbers $c$ such that $c=3, c=-1, f(c)=-1$ or $f(f(c))=-1$, or $f(f(f(c)))=-1$. The equation $f(x)=-1$ has exactly one root $x=1$. Thus, the last three equations are equivalent to $c=1, f(c)=1$, and $f(f(c))=1$. $f(c)=1$ has two solutions, $c=1 \pm \sqrt{2}$, and for each of these two values $c$ there are two preimages. It follows that the answer is $1+1+1+2+4=9$.