Let $f=\frac{1}{1+z^2}$.
I am trying to determine whether f is differentiable.
$f=\frac{1}{1+z^2}=\frac{1}{1+(x+iy)(x+iy)}=\frac{1}{1+x^2-y^2+2ixy}=\frac{1+x^2-y^2-2ixy}{(1+x^2-y^2)^2-4x^2y^2}$
Then I conclude that $u = \frac{1+x^2-y^2}{(1+x^2-y^2)^2-4x^2y^2}$ and $v=\frac{-2xy}{(1+x^2-y^2)^2-4x^2y^2}$.
For f to be differentiable $u_x$ should be equal to $v_y$ per Cauchy-Riemann equations. Using various calculators online I find that this is not the case.
$u_x=\frac{-2x^5+4x^3y^2-4x^3-10xy^4+12xy^2-2x}{\left(x^4-6x^2y^2+2x^2+y^4-2y^2+1\right)^2}$
$v_y=-\frac{2x\left(-3y^4+6x^2y^2+2y^2+x^4+2x^2+1\right)}{\left(x^4-6x^2y^2+2x^2+y^4-2y^2+1\right)^2}$
However another way to look at Cauchy-Riemann is that $\frac{\partial{f}}{\partial{\overline{z}}}$ should be $0$ and this is indeed the case, because f is a function $z$ only. How do I reconcile these two results?
I get $$f=\frac{1+x^2-y^2-2ixy}{(1+x^2-y^2)^2+4x^2y^2}.$$ I suggest computing the derivatives of $$u=\frac{1+x^2-y^2}{(1+x^2-y^2)^2+4x^2y^2}$$ and $$v=\frac{-2xy}{(1+x^2-y^2)^2+4x^2y^2}.$$