I have to show that $f(z) =\sin (\bar z)$ is not analytic anywhere.
One way is to check the CR equations by letting $\sin (z) = \sin(x+iy)$ and do some algebra. From the CR equations I obtained that $u_x = v_y$ occurs when $\cos(x) = 0$ and $\cosh(y)=0$. On the other hand, $v_x = -u_y$ occurs when $\sin(x)=0$ and $\sinh(y)=0$. The intersection of both sets of solutions is the empty set and thus I conclude $f(z)$ is not analytic anywhere.
But I figured why not try the Wirtinger derivatives since the function clearly is defined in terms of $\bar z$. So doing the partial derivative with respect to $\bar z$ I obtain the expression $\frac{\partial f}{\partial \bar z} = \cos(\bar z)$, which, while not identically zero, can be zero at certain values of $\bar z$.
My question is, I know that $\frac{\partial f}{\partial \bar z} = 0$ if and only if $f$ is analytic, but what if $\frac{\partial f}{\partial \bar z}$ is only zero at certain points? Does it mean it is not analytic anywhere? Does it mean it is analytic at those points? Does it even matter that $\frac{\partial f}{\partial \bar z}$ is an expression that can sometimes be zero?
I did consider using examples with singular points, like $f(z)=\frac{1}{z}$ to try and figure out how $\frac{\partial f}{\partial \bar z}$ looks like for a function that is analytic somewhere but not everywhere but in this case $\frac{\partial f}{\partial \bar z}=0$, and one example is never a deal breaker.
I know it doesn't really make sense to talk about possible analyticity on a point since a point is not an open set, but really I can't understand it.
Best Answer
$\newcommand{\dd}{\partial}$The map $z\mapsto\sin\bar z$ is holomorphic (at a point) sometimes. You should know that the Wirtinger derivative in the conjugate $\bar z$ is zero at a point $z_0$ if and only if the Cauchy-Riemann equations hold at that point $z_0$ (take real and imaginary parts on the definition).
Let's use the Wirtinger derivatives from their definition:
Application to $(x+iy)\overset{\sin\bar z}{\mapsto}\sin(x-iy)$ gives: $$\begin{align}\frac{\dd}{\dd\bar z}\sin\bar z&=\frac{1}{2}\left(\frac{\dd}{\dd x}\sin(x-iy)+i\frac{\dd}{\dd y}\sin(x-iy)\right)\\&=\frac{1}{2}\left(\cos(x-iy)+i(-i)\cos(x-iy)\right)\\&=\cos(x-iy)\end{align}$$
Exactly as you say, which equals zero if and only if $z=x+iy=\pi(n+1/2),\,n\in\Bbb Z$. What about the Cauchy Riemann equations? Let's be careful with what the equations imply.
Can we reconcile the two? Yes, we can. It seems as if all your calculations were correct, but your deductions had a mistake. First of all, $0\in\{\cos(x),\cosh(y)\}$ iff. $\cos(x)=0$ since $\cosh$ is nonzero on the real line. So the C-R criterion for differentiability (which is in fact identical to the Wirtinger criterion) tells us that our function is differentiable iff. $\cos(x)=0$ and one or both of $\sin(x),\sinh(y)$ are zero. However, $\sin^2(x)=1-\cos^2(x)=1$ if $\cos(x)=0$, so we have differentiability iff. $\cos(x)=0$ and $\sinh(y)=0$. The $\sinh$ function has only one real root, namely at $y=0$. Then, CR is stating that $y=0$ and $x$ is such that $\cos(x)=0$, so $z=\pi(n+1/2),\,n\in\Bbb Z$ exactly as the Wirtinger derivative criterion (much more efficiently) computed.
There are no contradictions here. More generally, if $f$ is holomorphic then $z\mapsto f(\bar z)$ is holomorphic at $z_0$ iff. $f'(z_0)=0$ (this is consistent with our example - at those mentioned $z$, $(\sin z)'=\cos(z)=0$). Try to prove this using the Wirtinger criterion.