Proving that a function derived from $\arctan(x/y)$ is continuous on $y\ne0$ $\Omega_1 = \{y > 0\} $
$\Omega_2 = \{y < 0\} $
$\Omega_3 = \mathbb R^2 \backslash \{x \leq 0 \ \ ; \  \ y = 0 \} $
\begin{equation}
  f(x,y) = \left \{
  \begin{aligned}
    &- \arctan \frac x y + \pi, && \text{if}\ (x,y) \in \Omega_1 \\
    & \frac {\pi} 2, && \text{if}  \ y = 0 \ ; \ x > 0 \\
    &- \arctan \frac x y, &&  \text{if}\ (x,y) \in \Omega_2
  \end{aligned} \right.
\end{equation} 

Prove that :$$ f \in C^1( \Omega_3) $$

So the obviously only problem is on the boundary of $\Omega_1$ and $\Omega_2$. I don't know how to prove neither continuity nor differentiability. Can you help me ? 
 A: Ok well, I have something of which I'm not completly sure. Do not hesitate to comment my proof.
What I'm going to show is that the $f$  function is differentiable, meaning it has continuous partial derivatives. This will give continuity.
First : (I note $f_x$ the partial derivative according to the first variable.) 
$$ g(x,y) := f_x(x,y) = \frac {y}{x^2+y^2}  $$
$$ h(x,y) := f_y(x,y) = -\frac {x}{x^2+y^2}  $$
So, let's take a look at those two functions.
We fixe $A(x,y)$ ; $B( x, \overline{y})   $ ; $ C( \overline{x} , \overline{y})$ 

We want to prove that :
$$\lim_{ (x,y) → (\overline{x} , \overline{y}) } g(x,y) = g(\overline{x} , \overline{y}) $$
Nevertheless :
$$|g(x,y) - g(\overline{x} , \overline{y})| \leq |g(x,y) - g(x , \overline{y})| + |g(x , \overline{y}) -g(\overline{x} , \overline{y}) | $$
and :
$$|g(x,y) - g(x , \overline{y})| = | \frac {y}{x^2 + y^2}| →_{(x,y) → (x , \overline{y})}   0$$
$$|g(x , \overline{y}) -g(\overline{x} , \overline{y}) | = 0$$
Then $g$ is continuous on $\Omega_3$.
We do the same thing for $h$ and this will conclude the proof.

We fixe another 3 points : $A(x,y)$ ; $B( \overline{x}, y)   $ ; $ C( \overline{x} , \overline{y})$ 
We want to prove that :
$$\lim_{ (x,y) → (\overline{x} , \overline{y}) } h(x,y) = h(\overline{x} , \overline{y}) $$
Nevertheless :
$$|h(x,y) - h(\overline{x} , \overline{y})| \leq |h(x,y) - h(\overline{x} , y)| + |h(\overline{x} , y) -h(\overline{x} , \overline{y}) | $$
and :
$$|h(x,y) - h(\overline{x} , y)|  →_{(x,y) → (\overline{x} , y)}   0$$
$$|h(\overline{x} , y) -h(\overline{x} , \overline{y}) |  →_{(\overline{x} , y) → (\overline{x} , \overline{y})}   0$$
Then $h$ is continuous on $\Omega_3$.
A: To see what's going on, let's consider a circle centered at the origin, and trace what happens to $f$ as we go around. Parametrize the circle in the standard way as $(x,y)=(\cos t,\sin t)$ for $-\pi < t\le \pi$ (That means it has radius $1$).
For $t=0$, we get $x=1,y=0$. That's the middle case of the definition of $f$, so $f(x(0),y(0))=\frac{\pi}{2}$.
For $0<t<\pi$, we have $y=\sin t>0$ and $f(x(t),y(t)) = \pi -\arctan\frac{\cos t}{\sin t} = \pi - (\frac{\pi}{2}-t) = \frac{\pi}{2}+t$. We used $\cot t =\tan(\frac{\pi}{2}-t)$ there.
For $-\pi < t < 0$, we have $y=\sin t < 0$ and $f(x(t),y(t))=-\arctan\frac{\cos t}{\sin t} = -(-\frac{\pi}{2}-t)=\frac{\pi}{2}+t$. It's still true that $\cot t=\tan(\frac{\pi}{2}-t)$, but since $\arctan$ takes values in $(-\frac{\pi}{2},\frac{\pi}{2})$, we use a different branch instead.
For $t=\pi$, we get $x=-1,y=0$. We can't fill in the value in a continuous way - it would have to be both $\frac{3\pi}{2}$ approaching from positive $t$ and $-\frac{\pi}{2}$ approaching from negative $t$ - but since the point isn't in the domain of $f$, we don't need to.
So then, on this nearly complete circle, $f(\cos t,\sin t) = \frac{\pi}{2}+t$. That's continuous, differentiable, and more. How do we extend it to the whole domain? Polar form; we can write an arbitrary point in the domain as $(x(r,\theta),y(r,\theta))=(r\cos\theta,r\sin\theta)$ for $r>0$ and $-\pi<\theta<\pi$. Then $f(x(r,\theta),y(r,\theta)) = \frac{\pi}{2}+\theta$. Verify that the coordinate functions for the transformation to polar form and its inverse are smooth, and we have it.
Added material:
Moreover, we can find more formulas for $f$ on various other subsets of the region it's defined on. Any half-plane in the domain has a shifted version of the arctangent formula; the easiest to find is the one for the right half-plane $\Omega_0=\{(x,y): x>0\}$. We get
$$f(x,y) = \arctan\frac yx +\frac{\pi}{2}\text{ if } x>0$$
That formula agrees with the ones we already have. Every point in the domain of $f$ is in at least one of the open regions $\Omega_0,\Omega_1,\Omega_2$, and $f$ is clearly smooth (infinitely differentiable) on each of those regions.
