What are some neat ways to differentiate $\arctan(\frac yx)$? By choosing the branch cut to be the line $\gamma=\{it\in \Bbb C: t\le 0\}$, we can write the complex logarithm as $$\text{Log}(z) = \log(|z|) + i\arg(z),$$
where $\arg(z) \in (-\pi,\pi].$ As a function of $x$ and $y$ (in $z=x+iy)$, we write
$$
\arg(z) = \arctan\left(\frac yx\right).
$$

How do we rigorously justify the following equalities 
  $$\begin{align}
\partial_x \left(\arctan\left(\frac yx\right)\right) &= \frac{-y}{x^2+y^2}, \\
\partial_y \left(\arctan\left(\frac yx\right)\right) &= \frac{x}{x^2+y^2}?
\end{align}$$

I usually see the above expressions come up when one want to show that the Cauchy-Riemann equations holds for $f(z) = \text{Log}(z)$. We can blindly apply the chain rule the get the answer but that is not rigorous because for $x=0$ the fraction $\frac yx$ is undefined.
To make it rigorous, we can rather write $\arg(z) = A(x,y)$ where
$$
A(x,y) = \begin{cases} \arctan\left(\frac yx\right) &; x\ne 0 \\
\pi/2 &;x=0, y>0 \\
-\pi/2 &; x=0, y<0
\end{cases}
$$
and compute the derivatives of this function carefully case by case instead but I find this method to be troublesome. Is there a neat way to get the result without having to consider $x=0$ as a separated case?

With $\arg z$ or $\arg(x,y)$ one denotes the "polar angle" of the point $$z\in\dot{\mathbb C}:={\mathbb C}\setminus\{0\},\quad {\rm resp.}, \quad (x,y)\in\dot{\mathbb R}^2:={\mathbb R}^2\setminus\{(0,0)\}\ .$$ I have put polar angle into quotes because in fact $\arg$ gives this angle only up to an integer multiple of $2\pi$. But your formula for $\arg$ is wrong: As given it is valid only for $x>0$. In reality one has
$$\arg(x,y)=\left\{\eqalign{\arctan{y\over x}\qquad&(x>0)\cr
{\pi\over2}-\arctan {x\over y}\qquad&(y>0)\cr
\pi+\arctan{y\over x}\qquad&(x<0)\cr
-{\pi\over2}-\arctan{x\over y }\qquad&(y<0)\cr}\right.\tag{1}$$
(up to $2\pi{\mathbb Z}$). If you draw a figure you can easily check that the definitions agree in the intersection of the two half-planes $x>0$ and $y>0$, etc.
While $\arg(x,y)$ is only determined "up to $2\pi{\mathbb Z}$" its gradient $\nabla\arg$ is a well defined vector field in $\dot{\mathbb R}^2$. You can easily check that for all lines in $(1)$ we have
$$\nabla\arg(x,y)=\left({\partial\arg\over\partial x},\>{\partial\arg\over\partial y}\right)=\left({-y\over x^2+y^2}, \>{x\over x^2+y^2}\right)\qquad\bigl((x,y)\in\dot{\mathbb R}^2\bigr)\ .$$
