Complex differentiability and higher derivatives of given function I have a question that goes:
Is $f:\mathbb{C} \rightarrow \mathbb{C}$ given by $f(z)=z^2+z|z|^2$ differentiable at $z=0$? If yes, find $f'(0).$ Also, does $f^{(n)}(0)$ exist for $n \geq 2?$
So what I did was try to express $f(z)= u(x,y)+iv(x,y)$ and then try the Cauchy-Reimann conditions and check whether the partial derivates $u_x, v_x, u_y, v_y$ are continuous or not, and if $u_x=v_y$ as well as $u_y=-v_x$. What I found was $$f(z)=x^2-y^2+x^3+xy^2+i(2xy+x^2y+y^3)$$
For this, I found that the partials are continuous at $z=0$ and the C-R equations are also satisfied. I then found $$f'(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}=2x+3x^2+y^2+i(2y+2xy)$$
I then tried to take to repeat the above method for $f'(z)$ to see if $f''(z)$ exists or not, and found that the C-R conditions are satisfied at $z=0$. But then I found $f''(z)$ and apply the same condition to check if the third order derivative exists or not and found that the C-R conditions are indeed not satisfied. So is the process for this part and my conclusion correct?

(This has been covered in comments, but the question should have an Answer...)
There's an error fairly early, in the parts you didn't show us: In fact $f$ verifies the C-R equations at the origin, but not at any other point.
Since $f'(z)$ does not exist for $z\ne0$ it's clear that $f''(0)$ cannot exist, since $$f''(0)=\lim_{z\to0}\frac{f'(z)-f'(0)}z.$$
That's so clear that bringing in the C-R equations for $f'$ seems silly. But just to clarify: since $u$ and $v$ are already taken, let's say $f'=U+iV$. Now $U$ and $V$ are undefined except at the origin, so there's no such thing as $U_x(0)$, as above. (Yes, if the C-R equations hold at one point then the function is differentiable at one point. BUt if the function  is defined at only one point the C-R equations cannot hold, becausethe real and imaginary parts cannot have partial derivatives...)
