# points where $f'(z)$ exists

I am revising complex analysis from scratch, please just tell am I right?

$f(z)=u_{xy}+iv_{xy}$, so by cauchy riemann I just need to solve the equation $u_x=v_y$ and $v_x=-u_y$ and those solutions are the points where $f'(z)$ exists. please confirm me.

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$u(x,y)=yx$, $v(x,y)=y^2$. Thus, $$\partial_xu=y\\ \partial_yu=x\\ \partial_xv=0\\ \partial_yv=2y$$ The partials are continuous everywhere. The Cauchy Riemann equations give: $$\partial_xu=\partial_yv\Leftrightarrow y=2y\Leftrightarrow y=0 \\ \partial_yu=-\partial_xv\Leftrightarrow x=0$$ $f$ is only complex differentiable at $(x,y)=0$ and $f^{\prime}(z)=0$
For b: $u(x,y)=x^2$, $v(x,y)=y^2$. Thus, $$\partial_xu=2x\\ \partial_yu=0\\ \partial_xv=0\\ \partial_yv=2y$$ The partials are continuous everywhere. The Cauchy Riemann equations give: $$\partial_xu=\partial_yv\Leftrightarrow 2x=2y\Leftrightarrow x=y \\ \partial_yu=-\partial_xv\Leftrightarrow 0=0$$ $f$ is only complex differentiable at $S=\left\{(x,x)\in \mathbb{C}:x\in \mathbb{R}\right\}$