Let y=f(x)=(x+2)/(5x-1),
y(5x-1)=x+2,
5xy-y=x+2,
5xy-x=y+2,
x(5y-1)=y+2,
x=(y+2)/(5y-1).
Let x=g(y)=(y+2)/(5y-1), so g(x)=(x+2)/(5x-1)=f(x).
But g(x)=f-1(x) so f-1(x)=f(x).
To check this, remember that f(f-1(x))=x by definition of the inverse.
f(f-1(x))=(f-1(x)+2)/(5f-1(x)-1)=((x+2)/(5x-1)+2)/(5(x+2)/(5x-1)-1).
Multiply top and bottom by 5x-1:
(x+2+10x-2)/(5x+10-5x+1)=11x/11=x,
which proves that the inverse of f(x)=(x+2)/(5x-1) is the same as f(x).
Also, f-1(f(x))=x since the function and its inverse are identical.