jmc

Expanding $f(f(x))=f(x)$ gives us $$(x^2-3x{)}^2-3(x^2-3x)=x^2-3x.$$Rather than expanding, we can subtract $x^2-3x$ from both sides to get $$(x^2-3x{)}^2-4(x^2-3x)=0.$$Factoring out $x^2-3x$ gives $(x^2-3x)(x^2-3x-4)=0$. Factoring each quadratic separately, we get $$x(x-3)(x+1)(x-4)=0.$$Thus the values of $x$ are $\boxed{0,3,-1,4}$.