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We have\begin{align*} f'(1) &= \lim_{z\to z_0}\frac{|z|^2-1}{z-1} = \lim_{(x,y)\to(1,0)}\frac{x^2+y^2-1}{(x-1) + iy} = \lim_{(x,y)\to (1,0)}\frac{(x^2+y^2-1)((x-1)-iy)}{(x-1)^2+y^2} \end{align*} Now let $g(x,y) =\frac{(x^2+y^2-1)((x-1)-iy)}{(x-1)^2+y^2}$.

$g(x,y)\rightarrow x+1$ as $(x,y)\rightarrow (1,0)$ along x-axis.

$g(x,y)\rightarrow (y+1)-(y+1)i$ as $(x,y)\rightarrow (1,0)$ along $x=y+1$ axis.

Is that enough for me to conclude $f'(1)$ does not exist.

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    You could also say: $$\frac{|1+h|^2-1}{h}=\frac{|h|^2+2\Re(h)}{h}$$ So if the limit as $h\to 0$ existed, then $\lim_{h\to 0}\frac{2\Re(h)}{h}=\lim_{h\to 0}\left[\frac{|1+h|^2-1}{h}-\frac{|h|^2}{h}\right]$ would also exist, which is clearly wrong – 2017-02-13
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    Yes. If the limit existed then its value would be path-independent. Here you get two different values for two different paths: $2$ ($x+1$ evaluated at $x=1$) and $1-i$ ($(y+1)-(y+1)i$ evaluated at $y=0$). – 2017-02-13

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Lets show that $$\lim_{z\to z_0 } \frac{|z|^2 - |z_0|^2}{z-z_0 } \qquad (\star)$$ does not exist where $z_0=x_0 + \mathrm i y_0\in \mathbf C \setminus \{0\}$. Therefore let us pick sequences $(z_n)_n \subset \mathbf C$ with $z_n=x_0 +\mathrm i y_n$ and $y_n \to y_0$ and $(z_n')_n \subset \mathbf C$ with $z_n'=x_n +\mathrm i y_0$ and $x_n \to x_0$. Then we see that $$\frac{|z_n|^2 - |z_0|^2}{z_n-z_0 } = x_0 - \mathrm i y_n - z_0 \to -2\mathrm i y_0 \qquad (n\to \infty)$$ and $$\frac{|z_n'|^2 - |z_0|^2}{z_n'-z_0 } = x_n - \mathrm i y_0 + z_0 \to 2x_0 \qquad (n\to \infty).$$ Since $-2\mathrm i y_0 \neq 2x_0$ we have proofed that the limit $(\star)$ does not exist because the limit $(\star)$ needs to hold for all sequences $(w_n)_n$ with $w_n \to z_0$. For $z_0=0$ it is easy to see that $(\star)$ exists.