Infinite fraction's derivative 
$f(x)=x+\dfrac{1}{x+\dfrac{1}{x+\dfrac{1}{x+\ldots}}}$
$f'\left(\dfrac{3}{2}\right)=?$

I tried to make equation like $y^2=xy+1$ but I can't made it clearly.

Using implicit differentiation,
$f^2-xf-1=0
$
gives
$0
=2ff'-f-xf'
=(2f-x)f'-f
$
so
$f'
=\dfrac{f}{2f-x}
$.
Since
$f(x)
=\dfrac{x+\sqrt{x^2+4}}{2}
$
(need the positive root),
$f(\frac32)
=\dfrac{\frac32+\sqrt{\frac94+4}}{2}
=\dfrac{\frac32+\sqrt{\frac{25}{4}}}{2}
=\dfrac{\frac32+\frac52}{2}
=2
$
so
$f'(\frac32)
=\dfrac{2}{4-\frac32}
=\frac45
$.
