<s>[INST] Find a second order Taylor polynomial for a function $z=\arctan\left( \frac{1-x+y}{1+x-y} \right)$ First we make substitution, so that $\arctan\left(\frac{1-u}{1+u} \right)$, where $u=x-y$. Then $u_x=1; u_y=-1$ Then  $z_u$ from there is expressed as $$z_u=\frac{1}{1+\left(\frac{1-u}{1+u}  \right)^2}=-\frac{1}{1+u^2}$$
Then $z_x=z_u \cdot u_x=z_u$ and $z_y=z_u \cdot u_y=-z_u$. Then it is noted, that because $z$ can be written at neighbourhood of $\left[ \begin{matrix} 0 \\ 0 \end{matrix}\right], \text{ then } z=\frac{\pi}{4}-x+y$. Why is it so? 
$z_{xx}=z_{uu} \cdot u_x, z_{yy}=z_{uu} \cdot u_y, z_{xy}=z_{uu} \cdot u_y$
$$z_{uu}=\frac{2}{(1+u^2)^2}$$
 [/INST] If you have
$$z=\arctan\left(\frac{1-u}{1+u} \right)$$ let $t=\frac{1-u}{1+u}$
$$\frac{dz}{du}=\frac{dz}{dt} \frac{dt}{du}=\frac1{1+t^2 } {\frac{-2}{(u+1)^2} }$$ and 
$$\frac1{1+t^2 }=\frac{(u+1)^2}{2 \left(u^2+1\right)}$$
$$\frac{dz}{du}=\frac{(u+1)^2}{2 \left(u^2+1\right)}{\frac{-2}{(u+1)^2} }=-\frac{1 } {u^2+1 }$$
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