I don't know what definitions you know, but I'll assume you know real calculus and linear algebra.
First, note that $z = x+iy = (1,1) \neq (1,-1) = x-iy =\overline{z}$, is direct the linear independence of $z$ with $\overline{z}$ (definition of linear algebra: no exists a scalar such that $\lambda z =\overline{z}$). Even, as vectors of $\mathbb{R}^{2}$ they form orthogonal directions.
By partial derivation of $f(x,y)=u(x,y)+iv(x,y)$, i.e.,
$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}$$
the above equality due to the uniqueness of the limit.
I define
\begin{equation*}
f'(z)=f'(x,y)=\frac{\partial u}{\partial x}(x,y)+i\frac{\partial v}{\partial x}(x,y)
\end{equation*}
and
\begin{equation*}
\begin{split}
\frac{\partial}{\partial z}&=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\\
\frac{\partial}{\partial \overline{z}}&=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)
\end{split}
\end{equation*}
it is possible to verify that $f'(z)=\frac{\partial f}{\partial z}$ and $\frac{\partial f}{\partial \overline{z}}=0$ by C-R.
The Cauchy - Riemann conditions hold if and only if $\frac{\partial f}{\partial\overline{z}}=0$
Proof:
($\Leftarrow$) Let $f:U\subseteq\mathbb{C}\to\mathbb{C}$, if $\frac{\partial f}{\partial\overline{z}}=0$, then:
\begin{equation*}
\frac{1}{2}\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right)=0
\end{equation*}
then (using $f= u+iv$:
\begin{eqnarray*}
0=\frac{\partial }{\partial x}(u+iv)+i\frac{\partial}{\partial y}(u+iv)
&=&\frac{\partial u}{\partial x}+i\frac{\partial u}{\partial y}+
i\frac{\partial v}{\partial x}-\frac{\partial v}{\partial y}\\
&=&\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+
i\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)
\end{eqnarray*}
Then (separating in real and imaginary parts):
$$\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)=0$$
and
$$\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)=0$$
concluding C-R.
($\Rightarrow$) By a construction of $f'(z)$, assuming C-R. Then:
\begin{eqnarray*}
f'(z)=f'(x,y)
&=&\frac{\partial u}{\partial x}(x,y)+i\frac{\partial v}{\partial x}(x,y)\\
&=&\frac{1}{2}\left(\frac{\partial u}{\partial x}(x,y)+i\frac{\partial v}{\partial x}(x,y)\right)+\frac{1}{2}\left(\frac{\partial u}{\partial x}(x,y)+i\frac{\partial v}{\partial x}(x,y)\right)
\\
&=&\frac{1}{2}\left(\frac{\partial u}{\partial x}(x,y)+i\frac{\partial v}{\partial x}(x,y)\right)+\frac{1}{2}\left(\frac{\partial v}{\partial y}(x,y)-i\frac{\partial u}{\partial y}(x,y)\right)
\\
&=&\frac{1}{2}\left(\frac{\partial u}{\partial x}(x,y)+i\frac{\partial v}{\partial x}(x,y)\right)+\frac{1}{2}\left(\frac{\partial v}{\partial y}(x,y)-i\frac{\partial u}{\partial y}(x,y)\right)
\\
&=&\dots\text{ sorting out, check it
}\dots\\
&=&\frac{\partial f}{\partial z}+\frac{\partial f}{\partial \overline{z}}
\end{eqnarray*}
But $f'(z) = \frac{\partial f}{\partial z}$, then $\frac{\partial f}{\partial \overline{z}}=0$.
I recommend seeing
- Stein. Shakarchi-Complex analysis. An Introduction - Princeton University.
- J. Bak & D.Newman-Complex analysis (2010).
- Alexander Isaev-Twenty one lectures on complex analysis (2017).
- Walter Rudin-Real and complex analysis
