Functions

If

$f(x) = \dfrac{a}{x + 2}$,

solve for the value of $a$ such that $f(0) = f^{-1}(-3a)$.

My attempt:
$\begin{array}{|rcll|} \hline f(x) &=& \dfrac{a}{x + 2} \\\\ f(0) &=& \dfrac{a}{0 + 2} \\\\ \mathbf{f(0)} &=& \mathbf{\dfrac{a}{2}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline f(x) &=& \dfrac{a}{x+2} \\\\ x &=& \dfrac{a}{f^{-1}(x)+2} \\\\ x\Big(f^{-1}(x)+2\Big) &=& a \\\\ xf^{-1}(x)+2x &=& a \\\\ xf^{-1}(x) &=& a-2x \\\\ f^{-1}(x) &=& \dfrac{a-2x}{x} \quad | \quad x =-3a\\\\ f^{-1}(-3a) &=& \dfrac{a-2(-3a)}{-3a} \\\\ f^{-1}(-3a) &=& \dfrac{a+6a}{-3a} \\\\ f^{-1}(-3a) &=& \dfrac{7a}{-3a} \\\\ \mathbf{f^{-1}(-3a)} &=& \mathbf{-\dfrac{7}{3}} \\\\ \hline f(0) &=& f^{-1}(-3a) \\\\ \dfrac{a}{2} &=& -\dfrac{7}{3} \\\\ \mathbf{a} &=& \mathbf{-\dfrac{14}{3}} \\ \hline \end{array}$