Exercise A12.5 of Hubbards' Vector Calculus, Linear Algebra, and Differential Forms I have a question about the solution to part (a) of exercise A12.5 of Hubbards' Vector Calculus, Linear Algebra, and Differential Forms.
Here is the exercise: 

Let $f $ be the function $f
  \begin{pmatrix}
    x \\
    y \\  
  \end{pmatrix}=sgn(y) \sqrt{ \frac {-x+\sqrt{x^2+y^2}}{2}} 
$ where $sgn(y)$ is the sign of y (i.e.,  $+1$ when $y>0$ , $0$ when $y=0$ ,$-1$ when $y<0$  )
a. show that $f$ is continuously differentiable on the complement of the half line $y=0, x\le 0$.

I've included part of the solution on the student solution manual to this exercise below: 
To show that $f$ is continuously differentiable on the locus where $y=0,x>0$:  

In a neighborhood of a point $\begin{pmatrix}
    x_0 \\
    y_0 \\  
  \end{pmatrix}$satisfying $x_0>0 $ , $y_0=0$ , we can write
  $-x+\sqrt{x^2+y^2}=-x+x\sqrt{1+\frac{y^2}{x^2}}= -x+x(1+\frac{1}{2}\frac{y^2}{x^2}+o(\frac{y^2}{x^2}))$
       $    =\frac{y^2}{2x}+o(\frac{y^2}{x})$,
  and since $sgn(y)y^2$ is of class $C^1$, the function is of class $C^1$ on the half axis $y=0, x> 0$.

I don't understand the bold part of the solution. Could you explain why the fact that $sgn(y)y^2$ is of class $C^1$ means that the function $f$ is continuously differentiable on the half-axis $y=0, x>0$ ?   
(The book uses the following definition for little $o$ notation: 
a function $f$ is in $o(h)$ if $\lim \limits_{x \to 0}\frac{f(x)}{h(x)}=0.$
)
 A: Definitely strange. In fact, $f_y(x_0,y_0)$ does not exists for $x_0 > 0$, $y_0 = 0$:
First, ignore the irrelevant 2 and consider 
$$f(x,y) = \hbox{sgn}(y)\sqrt{\sqrt{x^2+y^2} - x}.$$
$$
\lim_{y\to 0}\frac{\sqrt{\sqrt{x_0^2+y^2} - x_0} - 0}{y - 0} = \lim_{y\to 0}\sqrt{\frac1{2x_0} + \frac{o(y^2/x_0)}{y^2}} = \frac1{\sqrt{2x_0}}
,$$
so the lateral limits of
$$
\lim_{y\to 0}\frac{\hbox{sgn}(y)\sqrt{\sqrt{x_0^2+y^2} - x_0} - 0}{y - 0}
$$
are different.
Using
$$
\sqrt{x_0^2+y^2} - x_0 =
\frac{(\sqrt{x_0^2+y^2} - x_0)(\sqrt{x_0^2+y^2} + x_0)}{\sqrt{x_0^2+y^2} + x_0} =
\frac{y^2}{\sqrt{x_0^2+y^2} + x_0}
$$
gives the same result.
A: In order to better understand what's going on here introduce polar coordinates for the moment. Then
$$x=r\cos\phi,\qquad\sqrt{x^2+y^2}=r$$
and therefore
$${1\over2}\bigl(\sqrt{x^2+y^2}-x\bigr)=r{1-\cos\phi\over2}=r\sin^2{\phi\over2}\ .$$
It follows that
$$\sqrt{{1\over2}\bigl(\sqrt{x^2+y^2}-x\bigr)}=\sqrt{r}\left|\sin{\phi\over2}\right|\ .$$
As $\ {\rm sgn}\bigl(\sin{\phi\over2}\bigr)={\rm sgn}(y)$ we therefore obtain
$$f(x,y)=\sqrt{r}\sin{\phi\over2}={\root 4\of {x^2+y^2}}\ \sin{{\rm Arg}(x,y)\over2}\ .$$
As
$$\nabla{\rm Arg}(x,y)=\left({-y\over x^2+y^2},{x\over x^2+y^2}\right)$$the function $f$ is even real analytic in the indicated domain.
