I am tasked with showing that $$f(x+iy)=\frac{(1+i)x^3-(1-i)y^3}{x^2+y^2}\text{ if }x+iy\neq0;\quad f(0)=0$$ satisfies the Cauchy-Riemann equations at $0$, despite not being differentiable at $0$. We have $$u(x,y)=\frac{x^3-y^3}{x^2+y^2},\quad v(x,y)=\frac{x^3+y^3}{x^2-y^2}$$
And evaluating $u_x,v_x,u_y,v_y$, they don't satisfy the equations... which is fine, because they don't have to work everywhere, just zero... however they're not even defined at zero!
In a hint we are told to consider $x\rightarrow0$ along $y=x$ and $y=0$, but making these substitutions of $y$ into the partial derivatives yields constants that still don't satisfy $u_x=v_y$ and $u_y=-v_x$... So I'm stuck as to how to procede, or even as to whether I've already done something wrong...
$u_x(0,0)=\lim_{h \to 0} \frac {u(h,0)-u(0,0)} h=\lim_{h \to 0} 1=1$. Compute the other four partial derivatives at the origin and verify that C-R equations are satisfied.
Now $\frac {f(x+ix)-f(0)} {x+ix}\to \frac i {1+i} \neq 1$ whereas the limit along $y=0$ is $1$. Hence $f$ is not differentiable at $0$.