jmc

The definition of $f$ lets us evaluate $f(0)$: $$f(0)=\frac{a}{0+2}=\frac{a}{2}.$$Therefore we want to find all possible $a$ for which $$\frac{a}{2}=f^{-1}(3a).$$This is equivalent to $$f\left(\frac{a}{2}\right)=3a.$$When we substitute $x=\frac{a}{2}$ into the definition of $f$ we get $$f\left(\frac{a}{2}\right)=\frac{a}{\frac{a}{2}+2}=\frac{2a}{a+4},$$so we are looking for all solutions $a$ to the equation $$\frac{2a}{a+4}=3a.$$Multiplying both sides by $a+4$, we get $2a=3a(a+4)=3a^2+12a$, so $$3a^2+10a=0.$$Then $a(3a+10)=0$, so $a=0$ or $a=-10/3$. If $a=0$, then $f(x)=0$ for all $x\ne -2$, which means that the inverse function $f^{-1}(x)$ isn't defined, so $a=\boxed{-\frac{10}{3}}$.