TITLE: Differentiability of $\frac{xy^2(x+iy)}{x^2+y^4}$ at 0
QUESTION [3 upvotes]: This function is defined to be 0 at z=0  and the value at the rest of the points is given by that formula
I tried to use the Cauchy Riemann equations.
Here, $u=\frac{x^2y^2}{x^2+y^4}$
$u_x=\frac{2y^6x}{(x^2+y^4)^2}$
I'm getting $u_x=\frac{0}{0}$ at x=0, y=0. Similarly, I calculated $v_y$ and that's also 0/0 at z=0.
I think I'm not doing it right. Please give me some hint.

REPLY [1 votes]: Let $z=x+iy.$ Then
$$\frac{f(z)-f(0)}{z-0} = \frac{xy^2}{x^2+y^4}.$$
On the $x$-axis, this is $0,$ while on the path $(x,\sqrt x)$ this equals $1/2.$ Thus the limit of this as $x+iy\to 0$ fails to exist, which shows the complex derivative $f'(0)$ fails to exist.
{"set_name": "stack_exchange", "score": 3, "question_id": 2946683}