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)?
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.