Using Cauchy-Riemann equations and continuity of the partial derivatives to decide if $f(z)$ is analytic I'm doing an introductory course to complex analysis, and I'm having some trouble getting an overview on when a function is analytic and when it is not.
I have a function $f(z)$ which is defined for all $z=a+bi$. The question I'm trying to answer is "is $f(z)$ analytic?"
Taking the partial derivatives of $f(z)$:

$\frac{\partial u}{\partial x}=\frac{{x\left (x^{3} + 3 \, x y^{2} + 2 \, y^{3}\right)} }{{\left(x^{2} +
y^{2}\right)}^{2}},\frac{\partial v}{\partial x}=\frac{{x\left(x^{3} + 3 \, x y^{2} - 2 \, y^{3}\right)} }{{\left(x^{2} +
y^{2}\right)}^{2}}$
$\frac{\partial u}{\partial y}=-\frac{{y\left(2 \, x^{3} + 3 \, x^{2} y + y^{3}\right)} }{{\left(x^{2}
+ y^{2}\right)}^{2}},\frac{\partial v}{\partial y}=\frac{{y\left(-2 \, x^{3} + 3 \, x^{2} y + y^{3}\right)} }{{\left(x^{2}
+ y^{2}\right)}^{2}}$

So, these partial derivatives are not continuous when $x^2+y^2=0$, so $f(z)$ is not analytic in all of $\mathbb{C}$.
So now I have to check $\mathbb{C}\setminus \{(0,0)\}$; I still have to check if there is some open set where $f(z)$ is analytic, right?
And is the following correct?


*

*Looking at the points where the partial derivatives are continuous: if I can find an open set $G$ where the Cauchy-Riemann equations hold, then $f(z)$ is analytic on $G$.


Or is it so that if $\frac{\partial u}{\partial x}==\frac{\partial v}{\partial y}$ does not hold, then $f(z)$ is not analytic?
I am confused as to whether to think about the Cauchy-Riemann equations as a system of equations that can be solved or as a relationship that can be "checked".
The derivatives are continuous everywhere except $(0,0)$, but it is no so that $\frac{\partial u}{\partial x}==\frac{\partial v}{\partial y}$, or am I supposed to solve $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and then if the solution to this is not an open set (e.g. the solution is a line) I can say $f(z)$ is not analytic?
Thank you.
 A: 
And is the following correct?

*

*Looking at the points where the partial derivatives are continuous: if I can find an open set $G$ where the Cauchy-Riemann equations hold, then $f(z)$ is analytic on $G$.


Yes, that is correct. Analyticity is a local property, and a real-differentiable function can be analytic in part of its domain, and non-analytic in other parts.

Or is it so that if $\frac{\partial u}{\partial x} \equiv \frac{\partial v}{\partial y}$ does not hold, then $f(z)$ is not analytic?

In general, you cannot directly conclude that $f$ is nowhere analytic just from the fact that $\frac{\partial u}{\partial x}$ and $\frac{\partial v}{\partial y}$ are not globally the same function. They might coincide on some nonempty open set, and the Cauchy-Riemann equations could hold there.
Here, however, the partial derivatives are rational functions, and if two rational functions coincide in a nonempty open set, they are  globally identical. So here, you can conclude that $f$ is nowhere analytic from the fact that the partial derivatives are not the same rational function.

I am confused as to whether to think about the Cauchy-Riemann equations as a system of equations that can be solved or as a relationship that can be "checked".

That depends. If you are given an $f$, then they are a relationship that is to be checked. If you are given an $u$, or a $v$, then they may be a system of equations to be solved (given $u$, find $v$ so that $u+iv$ is analytic). Analyticity is a very restrictive property, so in general, given a real-valued function $u$, you can't find an analytic function $f$ whose real part is $u$. A necessary condition is that $u$ is harmonic. That condition is also locally sufficient, that is, every real-valued harmonic function is locally the real part of a holomorphic function, but in general there need not exist a holomorphic function on the entire domain of $u$ whose real part is $u$.
