Complex function only differentiable in $y=x^2$ 
Find a function $F: \mathbb{C} \rightarrow \mathbb{C}$ that is differentiable in the parabola $y=x^2$ and not differentiable in the rest of the complex plane.

Let $F(x,y) = u(x,y) + i v(x,y)$. If $F$ was differentiable in $y=x^2$ then the Cauchy Riemann equations would hold in that set, and applying the chain rule:
$$\frac{\partial}{\partial x}u(x,x^2) = \frac{\partial}{\partial y}v(x,x^2) 
 \implies \frac{\partial u}{\partial x} = 2x\frac{\partial v}{\partial y}$$
$$\frac{\partial}{\partial y}u(x,x^2) = -\frac{\partial}{\partial x}v(x,x^2) 
 \implies 2x\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$
Outside the parabola we would have that the C.R. equations would not hold because they are a necessary condition for differentiability.
I don't know how to continue from here as the C.R. equations haven't thrown much light about the funtion $F$.
 A: Here's an answer:$$f(x+yi)=-7x^3-3xy^2+6xy+\left(-y^3-9y^2+3x^2y-3x^2\right)i.\tag1$$And here's how I got this answer:


*

*I wrote $y-x^2$ as $\displaystyle\frac{z-\overline z}{2i}-\left(\frac{z+\overline z}2\right)^2=\frac z{2i}-\frac{\overline z}{2i}-\frac{z^2}4-\frac{\overline z^2}4-\frac{z\overline z}2$.

*Then I found a “primitive” of this expression with respect to $\overline z$:$$\frac{z\overline z}{2i}-\frac{\overline z^2}{4i}-\frac{z^2\overline z}4-\frac{\overline z^3}{12}-\frac{z\overline z^2}4.$$

*Finally, I replaced $z$ with $x+yi$ in order to get $(1)$. (Actually, I've also multiplied everything by $12$, in order to get only integer coefficients.)


Try it with other sets. It always works.
As far as the Cauchy-Riemann equations are concerned, this method always gives you a function $f(x+yi)=u(x,y)+v(x,y)i$ such that $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$. Besides, the equality $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ is basically the curve that you are interested in (in this specific case, this equality is equivalent to $24(y-x^2)=0$).
A: If $F(x +iy) = u(x,y) + i v(x,y)$ with
$$
 \begin{matrix}
u_x(x, y) = x^2 & u_y(x, y) = 0 \\
v_y(x, y) = 0 & v_y(x, y) = y
\end{matrix}
$$
then the Cauchy-Riemann equations are satisfied exactly for the points on the parabola $y = x^2$. A simple example is
$$
 F(x +iy) = \frac{x^3}{3} + i \frac{y^2}{2} \, .
$$
