smc

$I:f(0)=1$ Let $b=0$. Our equation becomes $f(a)f(0)=f(a)$, so $f(0)=1$. Therefore $I$ is always true. $II:f(-a)=\frac{1}{f(a)}\text{ for all }a$ Let $b=-a$. Our equation becomes $f(a)f(-a)=f(0)=1⟶f(-a)=\frac{1}{f(a)}$. Therefore $II$ is always true. $III:f(a)=\sqrt[3]{f(3a)}\text{ for all }a$ First let $b=a$. We get $f(a)f(a)=f(2a)$. Now let $b=2a$, giving us $f(3a)=f(a)f(2a)=f(a{)}^3⟶f(a)=\sqrt[3]{f(3a)}$. Therefore $III$ is always true. $IV:f(b)>f(a)\text{ if }b>a$ This is false. Let $f(x)=2^{-x}$, for example. It satisfies the conditions but makes $IV$ false. Therefore $IV$ is not always true. Since $I,II,\text{ and }III$ are true, the answer is $\boxed{\text{I, II, and III only}}$.