jmc

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

If $x\le 1$ and $f(x)=-3$, then $2-x=-3$, which leads to $x=5$. But this value does not satisfy the condition $x\le 1$. If $x>1$ and $f(x)=-3$, then $2x-x^2=-3$, or $x^2-2x-3=0$. This equation factors as $(x-3)(x+1)=0$, so $x=3$ or $x=-1$. The only value that satisfies the condition $x>1$ is $x=3$, so $f^{-1}(-3)=3$.

Next, we compute $f^{-1}(0)$, which is the value of $x$ such that $f(x)=0$.

If $x\le 1$ and $f(x)=0$, then $2-x=0$, which leads to $x=2$. But this value does not satisfy the condition $x\le 1$. If $x>1$ and $f(x)=0$, then $2x-x^2=0$, or $x^2-2x=0$. This equation factors as $x(x-2)=0$, so $x=0$ or $x=2$. The only value that satisfies $x>1$ is $x=2$, so $f^{-1}(0)=2$.

Finally, we compute $f^{-1}(3)$, which is the value of $x$ such that $f(x)=3$.

If $x\le 1$ and $f(x)=3$, then $2-x=3$, which leads to $x=-1$. Note that this value satisfies the condition $x\le 1$. If $x>1$ and $f(x)=3$, then $2x-x^2=3$, or $x^2-2x+3=0$. This equation can be written as $(x-1{)}^2+2=0$, which clearly has no solutions, so $f^{-1}(3)=-1$.

Therefore, $f^{-1}(-3)+f^{-1}(0)+f^{-1}(3)=3+2+(-1)=\boxed{4}$.