Definition
A complex function $f:\Omega\rightarrow\Bbb C$ defined in a open set $\Omega$ of $\Bbb C$ is said derivable at $z_0$ if the limit $$ \frac{df}{dz}(z_0):=\lim_{h\rightarrow0}\frac{f(z_0+h)-f(z_0)}h $$ exist.
Now we observe that if $\phi$ is the natural homeomorphism between $\Bbb R^2$ and $\Bbb C$ then $$ f(z)=\dots=\operatorname{Re}{f(z)}+\operatorname{Im}{f(z)}\cdot i=\operatorname{Re}{f(\operatorname{Re}z+\operatorname{Im}z\cdot i)}+\operatorname{Im}{f(\operatorname{Re}z+\operatorname{Im}z\cdot i)}\cdot i=\\\operatorname{Re}{f\big(\phi(\operatorname{Re}z,\operatorname{Im}z)\big)}+\operatorname{Im}{f\big(\phi(\operatorname{Re}z,\operatorname{Im}z)\big)}\cdot i $$ for any $z\in\Omega$ so that putting $$ \begin{cases}u(x,y):=\operatorname{Re}{f\big(\phi(x,y)\big)}\\v(x,y):=\operatorname{Im}{f\big(\phi(x,y)\big)}\end{cases} $$ for any $x,y\in\phi^{-1}[\Omega]$ then we conclud that $$ f(z)=\dots=\operatorname{Re}{f\big(\phi(\operatorname{Re}z,\operatorname{Im}z)\big)}+\operatorname{Im}{f\big(\phi(\operatorname{Re}z,\operatorname{Im}z)\big)}\cdot i=\\ u(\operatorname{Re}z,\operatorname{Im}z)+v(\operatorname{Re}z,\operatorname{Im}z)\cdot i=u\big(\phi^{-1}(z)\big)+v\big(\phi^{-1}(z)\big)\cdot i $$ for any $z\in\Omega$ or equivaletely $$ \phi(x+y\cdot i)=f\big(\phi(x,y)\big)=u\Big(\phi^{-1}\big(\phi(x,y)\big)\Big)+v\Big(\phi^{-1}\big(\phi(x,y)\big)\Big)\cdot i=u(x,y)+v(x,y)\cdot i $$ for any $(x,y)\in\phi^{-1}[\Omega]$. Now if the above limit was not path dependent then it would be $$ \frac{\partial u}{\partial x}+\frac{\partial v}{\partial x}\cdot i=\lim_{h\rightarrow 0}\frac{\big(u(x+h,y)+v(x+h,y)\cdot i\big)-\big(u(x,y).v(x,y)\cdot i\big)}{h}=\\ \lim_{h\rightarrow 0}\frac{f\big((x+h)+y\cdot i\big)-f(x+y\cdot i)}h=\frac{df}{dz}=\lim_{h\rightarrow 0}\frac{f\big(x+(y+h)\cdot i\big)-f(x+y\cdot i)}{h\cdot i}= \\\lim_{h\rightarrow 0}\frac{\big(u(x,y+h)+v(x,y+h)\cdot i\big)-\big(u(x,y)+v(x,y)\cdot i\big)}{h\cdot i}=\frac{\partial v}{\partial y}+\frac{1}i\cdot\frac{\partial u}{\partial y}=\frac{\partial v}{\partial y}-\frac{\partial u}{\partial y}\cdot i $$ that would implies the Cauchy-Riemann identities $$ \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\,\,\,\text{and}\,\,\,\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial x} $$ but unfortunately I am not able to prove that the limit $\frac{df}{dz}(z_0)$ is uniquely determined. In particular I know that $\Bbb C$ is a hausdorff space so that any net $(h_{\lambda})_{\lambda\in\Lambda}$ has at least one limit but I am not able to understand why this implies that if $(h_{\lambda})_{\lambda\in\Lambda}$ and $(k_{\lambda})_{\lambda\in\Lambda}$ are two distinct nets converging to $0$ then it must be $$ \lim_{\lambda\in\Lambda}\frac{f(z+h_\lambda)-f(z)}{h_\lambda}=\lim_{\lambda\in\Lambda}\frac{f(z+k_\lambda)-f(z)}{k_\lambda} $$ provided that the two limt exist. Another possible solution is to define the function $$ \varphi(h):=\frac{f(z_0+h)-f(z_0)}h $$ and so use this relevant result but unfortunately the function $\varphi$ is not defined at $0$ and so it is not possible to use the linked theorem -e.g. the limit of the function $\frac x{|x|}$ does not exist when $x$ approaches to zero and it is not definet at zero!!! So could someone help me, please?
Definition
If $X$ and $Y$ are topological space then $y_0\in Y$ is the limit of a function $f:X\to Y$ as $x$ approaches at the limit point $x_0\in X$ if and only if any net $\nu:\Lambda\to X\setminus\{x_0\}$ converging to $x_0$ is such that $f\circ\nu$ converges to $y_0$.
Lemma
If $\nu:\Lambda\to X$ is a net with values in a hausdorff space then it can converges at most one point.
Proof. See here.
Theorem
A function $f: X\rightarrow Y$ between topological spaces has at least one limit at any limit point when $Y$ is hausdorff.
Proof. So if $y_1$ and $y_2$ are two distinct limit of the function $f$ as $x$ approaches at the limit point $x_0$ then the net $\big(f(x_\lambda)\big)_{\lambda\in\Lambda}$ converges to $y_1$ and $y_2$ for any net $(x_\lambda)_{\lambda\in\Lambda}$ so that by the preceding lemma the two points must be equal.
So we conclude that the limit of the function $\varphi$ as $h$ approaches to $0$ is unique and so the the complex derivative is well define.