jmc

The number $f^{-1}(0)$ is the value of $x$ such that $f(x)=0$. Since the function $f$ is defined piecewise, to find this value, we must consider both cases $x\le 3$ and $x>3$.

If $x\le 3$ and $f(x)=0$, then $3-x=0$, which leads to $x=3$. Note that this value satisfies the condition $x\le 3$. If $x>3$ and $f(x)=0$, then $-x^3+2x^2+3x=0$. This equation factors as $-x(x-3)(x+1)=0$, so $x=0$, $x=3$, or $x=-1$. But none of these values satisfies $x>3$, so the solution is $x=3$, which means $f^{-1}(0)=3$.

Now we compute $f^{-1}(6)$, which is the value of $x$ such that $f(x)=6$.

If $x\le 3$ and $f(x)=6$, then $3-x=6$, which leads to $x=-3$. Note that this value satisfies the condition $x\le 3$. If $x>3$ and $f(x)=6$, then $-x^3+2x^2+3x=6$, or $x^3-2x^2-3x+6=0$. This equation factors as $(x-2)(x^2-3)=0$, so $x=2$, $x=\sqrt{3}$, or $x=-\sqrt{3}$. But none of these values satisfies $x>3$, so the solution is $x=-3$, which means $f^{-1}(6)=-3$.

Therefore, $f^{-1}(0)+f^{-1}(6)=3+(-3)=\boxed{0}$.