Proof of Implicit Differentiation (showing a statement is true) 
Prove that if $x^2+y^2-2y\sqrt{1+x^2} = 0$, then $dy/dx = x/\sqrt{1+x^2}$.

Whilst I have implicitly differentiated in terms of x in order to derive that
$$dy/dx = (-x+2xy/\sqrt{1+x^2})/(y\sqrt{1+x^2}-1-x^2)$$
however I am unsure as to what my next steps are. I believe that I will need to rearrange the original supposition in order to achieve the proof however I do require some help as to how can I do this as I cannot see what would possibly cancel out.
Additionally, does anyone have any particularly handy methods/techniques to be able to complete proofs of this format of question more easily? I do understand that there  is not one singular technique that can be used when completing this proofs but is there any strategy that minimizes the number of dead-ends that I come across in my working?
Thanks

We find from the given relation $ \ x^2 + y^2 \ = \  2y\sqrt{1+x^2}  \ $  that
$$ 2x \ + \ 2y·\frac{dy}{dx} \ \ =  \ \ 2·\frac{dy}{dx}·\sqrt{1+x^2} \ + \ 2y·\frac12·\frac{1}{\sqrt{1+x^2}}·2x \ $$
$$ \Rightarrow \ \   ( \ y \ - \ \sqrt{1+x^2} \ ) ·\frac{dy}{dx} \ \ =  \ \     y· \frac{1}{\sqrt{1+x^2}}· x \ - \ x \ \ \Rightarrow \ \  \frac{dy}{dx} \ \ = \ \   \frac{-x \ + \  \frac{xy}{\sqrt{1+x^2}}}{y \ - \ \sqrt{1+x^2} } \ \ . $$
Replacing (at least for the present) the radical using the original equation $ ( \ \sqrt{1+x^2} \ = \ \frac{x^2 + y^2}{2y} \ ) \ $ , we obtain
$$  \frac{dy}{dx} \ \ = \ \   \frac{  xy·\frac{2y}{x^2 + y^2} \ - \ x}{y \ - \ \frac{x^2 + y^2}{2y} } \ \ = \ \  \frac{  x·\left( \  2y^2  \ - \ [x^2 + y^2] \ \right)}{y·(x^2 + y^2) \ - \ \frac{(x^2 + y^2)^2}{2y} } \ \ = \ \ \frac{  x· (    y^2  \ - \  x^2      )·2y}{2y^2·(x^2 + y^2) \ - \  (x^2 + y^2)^2  }  $$
$$ = \ \ \frac{  x· (    y^2  \ - \  x^2      )·2y}{(2y^2 \ - \ x^2 \ - \ y^2)·(x^2 + y^2)   } \ \ = \ \ \frac{  x· (    y^2  \ - \  x^2      )·2y}{( y^2 \ - \ x^2 )·(x^2 + y^2)   } \ \ = \ \ x· \frac{  2y}{  x^2 + y^2    } 
$$
$$= \ \ x· \frac{1}{ \sqrt{1+x^2}   } \ \ ,  $$
again using the original relation.  (So it appears there may be an error in your expression for the implicit derivative.)
