Answer
$f^{-1}(x)=\dfrac{3x-1}{2}$
Work Step by Step
$f(x)=\dfrac{2x+1}{3}$
Substitute $f(x)$ by $y$:
$y=\dfrac{2x+1}{3}$
Solve for $x$:
$3y=2x+1$
$2x+1=3y$
$2x=3y-1$
$x=\dfrac{3y-1}{2}$
Interchange $x$ and $y$:
$y=\dfrac{3x-1}{2}$
Substitute $y$ by $f^{-1}(x)$:
$f^{-1}(x)=\dfrac{3x-1}{2}$
