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This is probably a silly question but let me ask.

As it is well known for a general function $f:\mathbb R^2\to \mathbb R$ which posesses partial derivatives of second order it is not necessarily true that

$\frac{\partial}{\partial x}(\frac{\partial f}{\partial y})=\frac{\partial}{\partial y}(\frac{\partial f}{\partial x}) $ and one concrete example is $f(x,y)=\frac{xy^3}{x^2+y^2},$ for $(x,y)\neq (0,0)$ nad $0$ at the origin. My question is about formal complex partial derivatives

$\frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)$ $\frac{\partial}{\partial \bar z}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)$ in the complex plane $\mathbb C\cong \mathbb R^2 (x+iy\cong z)$.

Is it always true that

$\frac{\partial}{\partial z}(\frac{\partial f}{\partial \bar z})=\frac{\partial}{\partial \bar z}(\frac{\partial f}{\partial z}), $ when both sides make sense or there exists an example of the above type (of course it can not be too regular)?

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    Yes that is exactly what "formal complex derivative making sense" means.2012-10-26

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Willie essentially answered the question, but here is the explicit computation in case someone finds it amusing:
$ \frac{\partial }{\partial z}\frac{\partial f}{\partial \bar z} = \frac14 \Delta f +\frac{i}{4}\left(\frac{\partial }{\partial x}\frac{\partial f}{\partial y} -\frac{\partial }{\partial y}\frac{\partial f}{\partial x}\right) $ and $ \frac{\partial }{\partial \bar z}\frac{\partial f}{\partial z} = \frac14 \Delta f - \frac{i}{4}\left(\frac{\partial }{\partial x}\frac{\partial f}{\partial y} -\frac{\partial }{\partial y}\frac{\partial f}{\partial x}\right) $ This means that the $z$- and $\bar z$- derivatives commute if and only if $x$- and $y$- derivatives commute. For real-valued $f$ the failure of commutativity is expressed as the presence of an imaginary term in the formal partials.