Math Challenges

Solution:
\( \displaylines{y-5=\frac{2}{\sqrt{3x}}\\ \sqrt{3x}=\frac{2}{y-5}\\ 3x=\left(\frac{2}{y-5}\right)^2\\ x=\frac13\left(\frac{2}{y-5}\right)^2\\ f^{-1}\left(x\right)=\frac13\left(\frac{2}{y-5}\right)^2} \)