All I really know is the quotient rule and the chain rule, but this problem blew up and took several pages of my notebook and is still wrong. Is there a way to solve this in less than a page?
$$ z=\arctan\left(\frac{x+y}{1-xy}\right) $$
$$ \text{find: } z_{xx} $$
On
Use the Chain Rule, then ${~\mathrm d\arctan(u)=(1+u^2)^{-1}\,\mathrm du~}$, and the Quotient Rule .
$\qquad\begin{align}\dfrac{\partial z}{\partial x} &=\dfrac{\partial}{\partial x}\left[\arctan\left(\dfrac{(x+y)}{(1-xy)}\right)\right] \\[1ex] &=\left.\dfrac{\mathrm d \arctan(u)}{\mathrm d u}\right\rvert_{u:=(x+y)/(1-xy)}\cdot\dfrac{\partial}{\partial x}\left[\dfrac{(x+y)}{(1-xy)}\right]\\[1ex]&=\dfrac{1}{1+[(x+y)/(1-xy)]^2}\cdot\dfrac{(1-xy)+y(x+y)}{(1-xy)^2}\end{align}$
Which simplifies into a very neat $z_x=(1+x^2)^{-1}$, and so...
HINT
I would start with noticing that
\begin{align*} \arctan\left(\frac{x + y}{1 - xy}\right) & = \arctan(x) + \arctan(y) \end{align*}