Either $f$ or $\overline{f}$ is complex-differentiable I wrote individually the "proof" for this question, but I believe my proof is not entirely correct, I wish someone else could help me to point out the error.
For simplicity, I will write here the question.

Let $f=u+iv$ be such that both $u$ and $v$ are real-differentiable function at $z=z_{0}$. If $\lim_{z\rightarrow 0}\left|\dfrac{f(z+z_{0})-f(z_{0})}{z}\right|$ exists, then either $f$ or $\overline{f}$ is complex-differentiable at $z=z_{0}$.

"Proof":
Denote 
\begin{align*}
L=\lim_{z\rightarrow 0}\left|\dfrac{f(z+z_{0})-f(z_{0})}{z}\right|,
\end{align*}
we have 
\begin{align*}
L^{2}=\lim_{z\rightarrow 0}\dfrac{f(z+z_{0})-f(z_{0})}{z}\cdot\dfrac{\overline{f}(z+z_{0})-\overline{f}(z_{0})}{\overline{z}}
\end{align*}
realizing $z=x+i\cdot0$ and $z=0+iy$ respectively we get
\begin{align*}
\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial\overline{f}}{\partial x}=-\dfrac{\partial f}{\partial y}\cdot\dfrac{\partial\overline{f}}{\partial y},
\end{align*}
so
\begin{align*}
\left|\dfrac{\partial f}{\partial x}\right|^{2}=-\left|\dfrac{\partial f}{\partial y}\right|^{2},
\end{align*}
then we must have 
\begin{align*}
\dfrac{\partial f}{\partial x}=\dfrac{\partial f}{\partial y}=0,
\end{align*}
so $\dfrac{\partial f}{\partial\overline{z}}=\dfrac{\partial\overline{f}}{\partial\overline{z}}=0$, they are both differentiable at $z=z_{0}$ with value $0$.
 A: The equation
$$\dfrac{\partial f}{\partial x}\cdot\dfrac{\partial\overline{f}}{\partial x}=-\dfrac{\partial f}{\partial y}\cdot\dfrac{\partial\overline{f}}{\partial y}$$ is incorrect.  Indeed, the right side is supposed to be 
$$\lim_{y\rightarrow 0}\dfrac{f(iy+z_{0})-f(z_{0})}{iy}\cdot\dfrac{\overline{f}(iy+z_{0})-\overline{f}(z_{0})}{\overline{iy}}=\frac{1}{i}\lim_{y\rightarrow 0}\dfrac{f(iy+z_{0})-f(z_{0})}{y}\cdot\frac{1}{-i}\lim_{y\rightarrow 0}\dfrac{\overline{f}(iy+z_{0})-\overline{f}(z_{0})}{y}$$ which comes out to $\dfrac{\partial f}{\partial y}\cdot\dfrac{\partial\overline{f}}{\partial y}$ with no minus sign since $i\cdot (-i)=1$.
A: @Eric Wofsey has pointed out the error. Anyway, your idea of the proof can be proceeded as follows.
So we need only to show that 
\begin{align*}
\dfrac{\partial f}{\partial\overline{z}}\dfrac{\partial\overline{f}}{\partial\overline{z}}=0.
\end{align*}
After a little computation we get
\begin{align*}
\dfrac{\partial f}{\partial\overline{z}}\dfrac{\partial\overline{f}}{\partial\overline{z}}&=\dfrac{\partial f}{\partial x}\dfrac{\partial\overline{f}}{\partial x}-\dfrac{\partial f}{\partial y}\dfrac{\partial\overline{f}}{\partial y}-\dfrac{1}{i}\left(\dfrac{\partial f}{\partial x}\dfrac{\partial\overline{f}}{\partial y}+\dfrac{\partial f}{\partial y}\dfrac{\partial\overline{f}}{\partial x}\right)\\
&=-\dfrac{1}{i}\left(\dfrac{\partial f}{\partial x}\dfrac{\partial\overline{f}}{\partial y}+\dfrac{\partial f}{\partial y}\dfrac{\partial\overline{f}}{\partial x}\right),
\end{align*}
where we have used the result that $\dfrac{\partial f}{\partial x}\dfrac{\partial\overline{f}}{\partial x}=\dfrac{\partial f}{\partial y}\dfrac{\partial\overline{f}}{\partial y}$.
Now we proceed by taking one direction of $z\rightarrow 0$:
\begin{align*}
L^{2}&=\lim_{x\rightarrow 0}\dfrac{f\left(z_{0}+\dfrac{x}{\sqrt{2}}+i\dfrac{x}{\sqrt{2}}\right)-f(z_{0})}{\dfrac{x}{\sqrt{2}}+i\dfrac{x}{\sqrt{2}}}\cdot\dfrac{\overline{f}\left(z_{0}+\dfrac{x}{\sqrt{2}}+i\dfrac{x}{\sqrt{2}}\right)-\overline{f}(z_{0})}{\dfrac{x}{\sqrt{2}}-i\dfrac{x}{\sqrt{2}}}\\
&=\lim_{x\rightarrow 0}\dfrac{f\left(z_{0}+\dfrac{x}{\sqrt{2}}+i\dfrac{x}{\sqrt{2}}\right)-f(z_{0})}{x}\cdot\dfrac{\overline{f}\left(z_{0}+\dfrac{x}{\sqrt{2}}+i\dfrac{x}{\sqrt{2}}\right)-\overline{f}(z_{0})}{x}\\
&=\left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}\right)\cdot\left(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right)\times\left(\dfrac{\partial\overline{f}}{\partial x},\dfrac{\partial\overline{f}}{\partial y}\right)\cdot\left(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right)\\
&=\dfrac{1}{2}\left(\dfrac{\partial f}{\partial x}\dfrac{\partial\overline{f}}{\partial x}+\dfrac{\partial f}{\partial y}\dfrac{\partial\overline{f}}{\partial y}+\dfrac{\partial f}{\partial y}\dfrac{\partial\overline{f}}{\partial x}+\dfrac{\partial f}{\partial x}\dfrac{\partial\overline{f}}{\partial y}\right)\\
&=\dfrac{\partial f}{\partial x}\dfrac{\partial\overline{f}}{\partial x}+\dfrac{1}{2}\left(\dfrac{\partial f}{\partial y}\dfrac{\partial\overline{f}}{\partial x}+\dfrac{\partial f}{\partial x}\dfrac{\partial\overline{f}}{\partial y}\right),
\end{align*}
but from the result we already have 
\begin{align*}
L^{2}=\dfrac{\partial f}{\partial x}\dfrac{\partial\overline{f}}{\partial x},
\end{align*}
so
\begin{align*}
\dfrac{\partial f}{\partial y}\dfrac{\partial\overline{f}}{\partial x}+\dfrac{\partial f}{\partial x}\dfrac{\partial\overline{f}}{\partial y}=0,
\end{align*}
so $\dfrac{\partial f}{\partial\overline{z}}\dfrac{\partial\overline{f}}{\partial\overline{z}}=0$ is justified.
