I'll take $A\subseteq\mathbb{C}$ to be an open set. A function $f:A\rightarrow
\mathbb{C}$ is holomorphic in $A$ if for every $z_{0}\in A$ there exists the
limit
$$
\lim_{z\rightarrow z_{0}}\frac{f(z)-f(z_{0})}{z-z_{0}}=\ell\in\mathbb{C},
$$
which just says that the derivative $f^{\prime}(z_{0})$ exists in $\mathbb{C}%
$. Some books unfortunately require $f^{\prime}$ to be continuous, which just
muddles the waters. It is not needed, you get it for free.
Once you know that $f$ is holomorphic, you can prove that $f^{\prime}$ is
continuous and it is itself holomorphic, and then you can prove that there
exist derivatives of any order and are all holomorphic. To prove this you fix
an open ball $B$ in $A$ and prove that for every rectifiable close curve
$\gamma$ contained in $B$ you have that $\int_{\gamma}f(z)\,dz=0$. Note that
to make sense of the integral you only need $f$ to be continuous and not even
differentiability. Cauchy proved this theorem assuming that $f^{\prime}$ was
continuous (and this is why some books add this to the definition) but later
Goursat proved that the result continues to hold assuming that $f$ is only differentiable.
Once you have $\int_{\gamma}f(z)\,dz=0$ for every rectifiable close curve
$\gamma$ contained in $B$ you prove Cauchy's formula, that is, that for every
$z_{0}\in B$
\begin{equation}
f(z_{0})=\frac{1}{2\pi i}\int_{\partial B(z_{0},r)}\frac{f(z)}{z-z_{0}%
}dz\label{cauchy formula}%
\end{equation}
where the closed ball $\overline{B(z_{0},r)}$ is contained in $B$. The point
now is that the right-hand side is a function $h(z_{0})$ which is given by an
integral depending on a parameter and as a function of $z_{0}$ you have that
$\frac{1}{z-z_{0}}$ is as regular as you want. So you can use theorems on
differentiation under the integral sign to conclude that that the right-hand
side has derivatives of any order and they are all continuous. In turn the
same is true for the left-hand side. So now you know that $f$ has derivatives
of any order and they are all continuous. At this point, if you define
$u(x,y):=$ real part of $f(x+iy)$ and $v(x,y):=$immaginary part of $f(x+iy)$,
then $u$ and $v$ are $C^{\infty}$ because they are given by the composition of
the function $(x,y)\mapsto x+iy$ which is $C^{\infty}$ with the function $f$
which is also $C^{\infty}$. In particular you can use the chain rule and take
second derivatives of $u$ and $v$ and prove that the Cauchy-Riemann equations
are satisfied and that $u$ and $v$ is harmonic (the mixed derivatives cancel
out because $u$ and $v$ are $C^{\infty}$). Hope this answers your second questions.
Now the converse implication is not perfect. There is a nice review article of
Gray and Morris
. One good result is the following. Take $U\subseteq
\mathbb{R}^{2}$ be an open set and let $u:U\rightarrow\mathbb{R}$ and
$v:U\rightarrow\mathbb{R}$ be such that there exist the partial derivatives
$\partial_{x}u$ and $\partial_{y}u$ and $\partial_{x}v$ and $\partial_{y}v$.
Note that the existence of partial derivatives DOES not imply that $u$ and $v$
are differentiable in $U$. You also assume that $u$ and $v$ satisfy the
Cauchy-Riemann equations.
These hypotheses are not enough, since they do not even imply that $u$ and $v$
are continuous. The function
$$
f(z):=\left\{
\begin{array}
[c]{ll}%
\exp(-z^{-4}) & \text{if }z\neq0\\
0 & \text{if }z=0,
\end{array}
\right.
$$
does not have a derivative at $z=0$ but the corresponding functions $u$ and
$v$ satisfy the Cauchy-Riemann equations in $\mathbb{R}^{2}$.
So you need an extra hypothesis. There are several variants. One that I like
is that $u:U\rightarrow\mathbb{R}$ and $v:U\rightarrow\mathbb{R}$ admit
$\partial_{x}u$ and $\partial_{y}u$ and $\partial_{x}v$ and $\partial_{y}v$ in
$U$, $u$ and $v$ satisfy the Cauchy-Riemann equations, and that the function
$f(z):=u(x,y)+iv(x,y)$, where $z=x+iy$, is continuous in the domain
$A=\{z=x+iy:\,(x,y)\in U\}$. The idea of the proof is to to mollify the
functions $u$ and $v$, taking $u_{\varepsilon}=\varphi_{\varepsilon}\ast u$
and $v_{\varepsilon}=\varphi_{\varepsilon}\ast v$, where $\varphi
_{\varepsilon}$ is a nice kernel, and prove that the $C^{\infty}$ functions
$u_{\varepsilon}$ and $v_{\varepsilon}$ satisfy the Cauchy-Riemann equations.
Using that, you can prove that $\int_{\gamma}f_{\varepsilon}(z)\,dz=0$ for
every rectifiable close curve $\gamma$ contained in $B$, where $f_{\varepsilon
}(z):=u_{\varepsilon}(x,y)+iv_{\varepsilon}(x,y)$, where $z=x+iy$, and then
continue as before to prove that $f_{\varepsilon}$ is analytic. In turn you
get the Cauchy formula (1) for $f_{\varepsilon}$ and then since $f$ is
continuous, you can let $\varepsilon\rightarrow0$ to prove that $f$ satisfies
the Cauchy formula and so is holomorphic.
Hope this answers your first question. Read the paper, it is well-written.
Best Answer
There are a lot of equivalent characterizations of holomorphic functions, but the main part of the story is that for a given function $$ f: \mathbb{R}^2 \mapsto \mathbb{R}^2 $$ all you need to know is that $f$ is continuous. Then some equivalent conditions are:
$f$ is complex differentiable (the differential $d f$ exists and is complex linear),
$f$ is locally integrable (edit: I meant to say that $f$ has an antiderivative, this point and point 3 are usually called Morera's theorem),
integrals along closed paths of $f$ are zero,
$f$ is conformal,
Cauchy's integral formula holds for $f$,
$f$ is represented by a power series around every point,
the Cauchy-Riemann equations hold,
etc.
For some of these conditions to make sense you need to know that $f$ is once differentiable as a real function. But none are related to $f$ being smooth (infinitely often differentiable as a real function), so knowing that $f$ is smooth is knowledge that is orthogonal to knowing that $f$ is holomorphic.
Edit: Of course every holomorphic function is smooth. What I meant to say is that being smooth is not "closer to being holomorphic than simply being differentiable" in any sense. AFAIK, there is no statement like "a continuous function is holomorphic if it is smooth + some other condition".