This is my question:
Prove that the function $f(z) = u+iv$, where $$ f(z) = \begin{cases} \frac{x^3(1+i) - y^3(1-i)}{x^2+y^2}, &\text{ if }z\neq 0,\\ 0 &\text{if }z=0.\end{cases}$$ satisfies the Cauchy-Riemann equation at the origin, but the derivative of $f$ at $z=0$ does not exists.
I solved the second part about the derivative but was not able to prove the first part, that is, $u_x=v_y$ at $z=0$. Please kindly direct me how to solve it: I got
\begin{align*} u_x&=\frac{x^4 +3x^2x^2 + 2xy^3}{(x^2 + y^2)^2}, \\ u_y&=\frac{ -y^4 -2x^3y -3x^2y^2}{(x^2 + y^2)^2}, \\ v_x&= \frac{x^4 - 2xy^3 + 3x^2y^2}{(x^2 + y^2)^2}, \\ v_y&=\frac{y^4+2x^2y^2-2x^3y}{(x^2 + y^2)^2}. \end{align*}
and there remains the question of how to prove $u_x=v_y$ at $(x,y)=(0,0)$.