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

A simpler way: the identity
$$x^2+y^2-2y\sqrt{1+x^2} = 0$$
is equivalent to
$$y^2+(1+x^2)-2y\sqrt{1+x^2} = 1$$
that is
$$(y-\sqrt{1+x^2})^2=1$$
and differentiating  in terms of $x$ both sides, we obtain
$$2(y-\sqrt{1+x^2})\left(y'-\frac{x}{\sqrt{1+x^2}}\right)=0.$$
The first factor is always different from zero: in fact letting $y=\sqrt{1+x^2}$ into the given equation we get
$$x^2+(1+x^2)-2(1+x^2) = 0$$
that is $-1=0$ which is impossible. Therefore the second factor has to be zero, and we find
$$y'=\frac{x}{\sqrt{1+x^2}}.$$
Preliminary step. Notice that by the two dimensional implicit function theorem, $y'(x)$ exists (and therefore we are allowed to take the derivative in terms of $x$) if
$$\frac{\partial F}{\partial y}=2(y-\sqrt{1+x^2})\not=0$$
where $F(x,y)=x^2+y^2-2y\sqrt{1+x^2}$, which holds as shown above.
