btw, at the end of the proof of Morera, they say that $F’=f$. Now, I do see the equivalence with the fundamental thm of calculus for 1D, but.. how does it work in the complex case? Does that always hold? I actually just tried to calculate it directly;
$$
F’=\dfrac{\partial}{\partial z}F=1/2\left(\dfrac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)F,
$$
but that gave me $1/2(\operatorname{Re}f+i\operatorname{Im}f+\operatorname{Im}f-i\operatorname{Re}f\}$. Did I just make a mistake? And can I use some fund. thm of calculus directly?

When $y\geq x^2$ we have that $f(x,y)=y-x\geq x^2-x\geq -\frac{1}{4}$. Since $-\frac{1}{4}$ is smaller than $0$ (the value of the function when $y<x^2$) it follows that the function $f(x,y)$ has a minimum at $=\frac{1}{2}$ which is equal to $-\frac{1}{4}$.
Is this correct? If yes, could we improve the justification?

About the maximum: In the first case, $y$ is greater than $x^2$, so I think that the value of the function can grow infinitely, i.e. it has no maximum. Is this correct? But how could we justify that formally?