Statement in question this leads to an image but I, unfortunately, don't have enough reputation points.
This is questioning the following equality:
$$\mathrm df(z)=\mathrm du+i\mathrm dv=F(\mathrm dx+i\mathrm dy)=F~\mathrm dz$$
where $F$ is the derivative of $f(z)$.
This is my thinking: $\mathrm dz =\mathrm dx + i\mathrm dy$, therefore, the first three statements are correct, but the $F~\mathrm dz$ statement may not be. Concern is that the last statement is missing $()$. $F ~\mathrm dz ≠ F(\mathrm dz)$, does it?.
Where $F~\mathrm dz$ is the derivative of $f$ multiplied by $\mathrm dz$ and $F(\mathrm dz)$ is the derivative of $f$ in terms of $\mathrm dz$. I think the correct statement should be $f(z)~\mathrm dz$ or $F(\mathrm dz)$.
Thanks for reading. all the best
The "derivative of $f$ in terms of $dz$" doesn't even make sense.
$F$ is a function on the complex plane. It takes complex numbers as its arguments. But $dz$ is not a complex number. It is a differential form. So until you define how $F$ should act on these new objects (differential forms), you cannot evaluate $F$ at them.
That expression is ambiguous, and unfortunately you've chosen the wrong interpretation. By $F(dx + idy)$, it just means "the function $F$ multiplied by the differential $dx + idy$. That is, $F(z)(dx + idy)$. This holds by the two-dimensional form of the chain rule.
Where $z = x + iy$ and $f(z) = u(x,y) + iv(x,y)$, we have $$df = du + idv$$ (the differential operator is linear) and $$du = u_xdx + u_ydy\\dv = v_xdx + v_ydy$$ So $$df = (u_x + iv_x)dx + i (v_y - iu_y)dy $$ The Cauchy-Riemann conditions tell us that $u_x = v_y, u_y = -v_x$, so in fact $v_y - iu_y = u_x + iv_x$. Define $F(z) = F(x,y)$ to be the common value. $$df = F(z)dx + iF(z)dy = F(z)(dx + idy) = F(z)dz$$ Or hiding the $z$ dependency in $F$, just like it is already hidden in $df, dx, dy, dz$, and this is $$df = Fdz$$ From this it follows that $F = f'$.