Answer
$a)$ $f^{-1}(x)=\dfrac{7}{x+3}$
$b)$ $f(f^{-1}(x))=f^{-1}(f(x))=x$
Work Step by Step
$f(x)=\dfrac{7}{x}-3$
$a)$
Substitute $f(x)$ by $y$:
$y=\dfrac{7}{x}-3$
Interchange $x$ and $y$:
$x=\dfrac{7}{y}-3$
Solve for $y$:
$x+3=\dfrac{7}{y}$
$y(x+3)=7$
$y=\dfrac{7}{x+3}$
Substitute $y$ by $f^{-1}(x)$:
$f^{-1}(x)=\dfrac{7}{x+3}$
$b)$
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:
$f(f^{-1}(x))=\dfrac{7}{\Big(\dfrac{7}{x+3}\Big)}-3=\dfrac{7(x+3)}{7}-3=...$
$...=x+3-3=x$
Complete the verification process by finding $f^{-1}(f(x))$. Substitute $x$ by $f(x)$ in $f^{-1}(x)$ and simplify:
$f^{-1}(f(x))=\dfrac{7}{\dfrac{7}{x}-3+3}=\dfrac{7}{\Big(\dfrac{7}{x}\Big)}=\dfrac{7x}{7}=x$
Since $f(f^{-1}(x))=f^{-1}(f(x))=x$, the equation for the inverse function found is correct.