Answer
$a)$ $f^{-1}(x)=-\dfrac{x+3}{x-2}$
$b)$ $f(f^{-1}(x))=f^{-1}(f(x))=x$
Work Step by Step
$f(x)=\dfrac{2x-3}{x+1}$
$a)$
Substitute $f(x)$ by $y$:
$y=\dfrac{2x-3}{x+1}$
Interchange $x$ and $y$:
$x=\dfrac{2y-3}{y+1}$
Solve for $y$. Start by taking $y+1$ to multiply the left side:
$x(y+1)=2y-3$
$xy+x=2y-3$
Take $2y$ to the left side and $x$ to the right side:
$xy-2y=-x-3$
Take out common factor $y$ from the left side:
$y(x-2)=-x-3$
Take $x-2$ to divide the right side:
$y=\dfrac{-x-3}{x-2}$
$y=-\dfrac{x+3}{x-2}$
Substitute $y$ by $f^{-1}(x)$:
$f^{-1}(x)=-\dfrac{x+3}{x-2}$
$b)$
Verify the equation for the inverse function found by finding $f(f^{-1}(x))$. Substitute $x$ by $f^{1}(x)$ in $f(x)$ and simplify:
$f(f^{-1}(x))=\dfrac{2\Big(-\dfrac{x+3}{x-2}\Big)-3}{-\dfrac{x+3}{x-2}+1}=\dfrac{-\dfrac{2x+6}{x-2}-3}{\Big(\dfrac{x-2-x-3}{x-2}\Big)}=...$
$...=\dfrac{-\Big(\dfrac{2x+6+3x-6}{x-2}\Big)}{\Big(\dfrac{-5}{x-2}\Big)}=\dfrac{-\Big(\dfrac{5x}{x-2}\Big)}{-\Big(\dfrac{5}{x-2}\Big)}=x$
Complete the verification process by finding $f^{-1}(f(x))$. Substitute $x$ by $f(x)$ in $f^{-1}(x)$ and simplify:
$f^{-1}(f(x))=-\dfrac{\dfrac{2x-3}{x+1}+3}{\dfrac{2x-3}{x+1}-2}=-\dfrac{\Big(\dfrac{2x-3+3x+3}{x+1}\Big)}{\Big(\dfrac{2x-3-2x-2}{x+1}\Big)}=...$
$...=-\dfrac{\Big(\dfrac{5x}{x+1}\Big)}{\Big(\dfrac{-5}{x+1}\Big)}=x$
Since $f(f^{-1}(x))=f^{-1}(f(x))=x$, the equation for the inverse function found is correct.