These functions exists (and are invertible) as long as the denominators are not zero: so we want
[tex]x-a\neq 0 \iff x\neq a[/tex]
for [tex]f(x)[/tex], and
[tex]x-b\neq 0 \iff x\neq b[/tex]
for [tex]g(x)[/tex]
Now we show that they are inverses of each other: we want to show that
[tex]f(g(x))=g(f(x))=x[/tex]. Let's start with [tex]f(g(x))=x[/tex]. We have
[tex]f(g(x))=\dfrac{1}{g(x)-a}+b=\dfrac{1}{\frac{1}{x-b}+a-a}+b=\dfrac{1}{\frac{1}{x-b}}+b=x-b+b=x[/tex]
And in the very same way we show that [tex]g(f(x))=x[/tex]
