The function $f(x)$ satisfies
\[f(x + y) = f(x) f(y)\]for all real numbers $x$ and $y.$  Find all possible values of $f(0).$

Enter all the possible values, separated by commas.
Setting $x = y = 0,$ we get
\[f(0) = f(0)^2,\]so $f(0) = 0$ or $f(0) = 1.$  The constant functions $f(x) = 0$ and $f(x) = 1$ show that both $\boxed{0,1}$ are possible values of $f(x).$