Question:
If $f(x)=\dfrac{a}{x+2}$, solve for the value of $a$ so that $f(0)=f^{-1}(3a)$.

Answer:
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 \[\frac a2=f^{-1}(3a).\]This is equivalent to  \[f\left(\frac a2\right)=3a.\]When we substitute $x=\frac a2$ into the definition of $f$ we get  \[f\left(\frac a2\right)=\frac{a}{\frac a2+2}=\frac{2a}{a+4},\]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 $x \neq -2$, which means that the inverse function $f^{-1}(x)$ isn't defined, so $a = \boxed{-\frac{10}{3}}$.