jmc

Suppose the function $f(x)=0$ has only one distinct root. If $x_1$ is a root of $f(f(x))=0,$ then we must have $f(x_1)=r_1.$ But the equation $f(x)=r_1$ has at most two roots. Therefore, the equation $f(x)=0$ must have two distinct roots. Let them be $r_1$ and $r_2.$

Since $f(f(x))=0$ has three distinct roots, one of the equations $f(x)=r_1$ or $f(x)=r_2$ has one distinct root. Without loss generality, assume that $f(x)=r_1$ has one distinct root. Then $f(x)=x^2+6x+c=r_1$ has one root. This means $$x^2+6x+c-r_1$$must be equal to $(x+3{)}^2=x^2+6x+9=0,$ so $c-r_1=9.$ Hence, $r_1=c-9.$

Since $r_1$ is a root of $f(x)=0,$ $$(c-9{)}^2+6(c-9)+c=0.$$Expanding, we get $c^2-11c+27=0,$ so $$c=\frac{11\pm \sqrt{13}}{2}.$$If $c=\frac{11-\sqrt{13}}{2},$ then $r_1=c-9=-\frac{7+\sqrt{13}}{2}$ and $r_2=-6-r_1=\frac{-5+\sqrt{13}}{2},$ so $$f(x)=x^2+6x+\frac{11-\sqrt{13}}{2}=\left(x+\frac{7+\sqrt{13}}{2}\right)\left(x+\frac{5-\sqrt{13}}{2}\right)=(x+3{)}^2-\frac{7+\sqrt{13}}{2}.$$The equation $f(x)=r_1$ has a double root of $x=-3,$ and the equation $f(x)=r_2$ has two roots, so $f(f(x))=0$ has exactly three roots.

If $c=\frac{11+\sqrt{13}}{2},$ then $r_1=c-9=\frac{-7+\sqrt{13}}{2}$ and $r_2=-6-r_1=-\frac{5+\sqrt{13}}{2},$ and $$f(x)=x^2+6x+\frac{11+\sqrt{13}}{2}=\left(x+\frac{7-\sqrt{13}}{2}\right)\left(x+\frac{5+\sqrt{13}}{2}\right)=(x+3{)}^2+\frac{-7+\sqrt{13}}{2}.$$The equation $f(x)=r_1$ has a double root of $x=-3,$ but the equation $f(x)=r_2$ has no real roots, so $f(f(x))=0$ has exactly one root.

Therefore, $c=\boxed{\frac{11-\sqrt{13}}{2}}.$