Answer
$f^{-1}(x)=\frac{3x+1}{x-2}$
Work Step by Step
$f(x)=\frac{2x+1}{x-3}$
$y=\frac{2x+1}{x-3}$
a.
$x=\frac{2y+1}{y-3}$
$xy-3x=2y+1$
$xy-2y=3x+1$
$y(x-2)=3x+1$
$y=\frac{3x+1}{x-2}$
$f^{-1}(x)=\frac{3x+1}{x-2}$
b.
$f(f^{-1}(x))=\frac{2\left(\frac{3x+1}{x-2}\right)+1}{\frac{3x+1}{x-2}-3}=\frac{7x}{7}=x$
and
$f^{-1}(f(x))=\frac{3\left(\frac{2x+1}{x-3}\right)+1}{\frac{2x+1}{x-3}-2}=\frac{7x}{7}=x$
