[tex]\\ \sf\longmapsto f(x)=\dfrac{x-1}{x+1}[/tex]
[tex]\\ \sf\longmapsto f^{-1}(x)=\dfrac{1}{f(x)}=dfrac{x+1}{x-1}[/tex]
Now
[tex]\\ \sf\longmapsto f^{-1}(3)=\dfrac{3+1}{3-1}=4/2=2[/tex]
And
[tex]\\ \sf\longmapsto (f^{-1})'(3)=f(3)=\dfrac{3-1}{3+1}=2/4=1/2[/tex]
Done
