Not clear whether ratio of two holomorphic functions is holomorphic function except points whether denominator is zero By the link Proof that a function is holomorphic it is told

Elementary operations or compositions of holomorphic functions give holomorphic functions on the maximal domain where the functions are defined. This is a consequence of the rules of derivation for product, ratio and compositions of functions.

But per my understand I have counterexample
$$
f(z) = f(x + \imath y) = \frac{x - \imath y}{x + \imath y}.
$$
I calculate by dividing complex that
$$
f(x + \imath y) = 1 - 2\imath\frac{xy}{x^2 + y^2} = u(x, y) + \imath v(x, y).
$$
Then I verify Cauchy-Riemann criteria
$$
\frac{\partial u}{\partial x} = 0,
$$
and while this
$$
\frac{\partial v}{\partial y} = -2x\frac{x^2 - y^2}{(x^2 + y^2)^2},
$$
which means that $f$ is not holomorphic.
Did I made mistake in calculations or this means that cited statement is not correct?

$x-iy$ is not a holomorphic function. The Cauchy-Riemann equations are not satisfied: $\partial x/\partial x = 1$ while $\partial (-y)/\partial y = -1$. So the equation $\partial u/\partial x = \partial v / \partial y$ does not hold at any point $(x,y)$. So this function is not holomorphic, and so your example is not a counterexample.
