In "Micheal's E. Taylor's" Complex Analysis, i'm having trouble computing the Partial Derivatives, my initial approach leading up to where i'm stuck in is detailed in $(0)$. Specifically in $(4)$, i'm having trouble verifying my set-up of the partial derivatives is valid.
$(0)$
Let $\Omega \subset \mathbb{C}$, be open and assume $K \subset \Omega$ is compact, assume $f,g \in C(\Omega)$ and in $(1.2)$, the following assumption can be made:
$(0.1)$ $$\triangle_{h}f(z) \rightarrow f'(z), \triangle_{h}g(z) \rightarrow g'(z).$$, $$ \text{Remark} : (0.1), and \, (0.2)\,\text{are} \, \text{Uniformaly on K}$$ One has to show that in $(0.2)$
$(0.2)$ $$ \triangle_{h}(f(z)g(z)) = \triangle_{h}f'(z)g(z) + f(z)g'(z)$$
$$\text{Remark}$$
The $f(z)$, observed within $(0.1)$, is formally defined as the following in $(1.3)$, one can make subtle operations by exploiting the Lapalcian Opeator on the RHS side of $(1.)$
$(1.3)$
$$(1) \, \, \, \, \, \, \triangle{h}f(z) = \frac{1}{h}(f(z+h) - f(z))$$ $$(2) \, \, \, \, \, \, \nabla^2_{h}f(z) = \nabla^2_{h}(\frac{1}{h}(f(z+h) - f(z)))$$ $$(3) \, \, \, \, \, \, \nabla_{} \cdot \nabla_{h}f(z) =\nabla_{} \cdot \nabla_{h} (\frac{1}{h}(f(z+h) - f(z)))$$
$$\text{Lemma}:$$
To verify the Holomorphcity of $f$, one can use the Cauchy-Riemann Equations, formally developed: in $(1.4)$
$(1.4)$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $Let $u, v: \left\{ {\left({x, y}\right) \in \mathbb{R^2} }\, \middle\vert \, {x+iy = z \in D }\right\} \to \mathbb{R}$ be two real-valued functions:
$$u(x,y) = Re(f(z))$$ $$u(x,y) = Im(f(z))$$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $The $f$ is complex-differentiable in $\text{D}$ if and only if: $$(1): \quad {\partial u}/{\partial x} = {\partial v}/{\partial y}$$ $$(2): \quad {\partial u}/{\partial y} = - {\partial v}/{\partial x}$$
If the conditions are true then: $$f' \left({z}\right) = {\partial f}/{\partial x} \left({z}\right) = -i {\partial f}/{\partial y} \left({z}\right)$$
Applying our recent development in $(1.4)$, first we can denote the real and complex parts of our function in $(3.)$ as follows:
$(3.)$ $$ \nabla_{} \cdot \nabla_{h}f(v+iy)=\nabla_{} \cdot \nabla_{h} (\frac{1}{h}(f(z)+h) - f(z))$$
$$\nabla_{} \cdot \nabla_{h}f(z)=\nabla_{} \cdot \nabla_{h} (\frac{1}{h}(f(z)+h) - f(z))$$
$$\text{Remark}$$: $ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $The Real and Complex Parts of our function can be denoted as follows: $$\nabla_{} \cdot \nabla_{h}(\text{Re}(f(z))$$
$$\nabla_{} \cdot \nabla_{h}(\text{Im}(f(z))$$
$$\text{Lemma}$$
$ \, \, \, \, \, \, \, \, \, \, \, $$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $Applying,$(1.)$ and $(2.)$ one should be able to observe the following in $(4)$
$(4)$
$$ \partial_{u}(\nabla_{} \cdot \nabla_{h}f(v+iy))= \partial_{v}(\nabla_{} \cdot \nabla_{h}f(v+iy))$$
$$ \partial_{u}(\nabla_{} \cdot \nabla_{h}f(v+iy))= -\partial_{v}(\nabla_{} \cdot \nabla_{h}f(v+iy))$$
$$\partial_{u}(\nabla_{} \cdot \nabla_{h} (\frac{1}{h}(f(u+iy)+h) - f(u+iy))) = \partial_{v}(\nabla_{} \cdot \nabla_{h} (\frac{1}{h}(f(v+iy)+h) - f(v+iy))) $$
$$\partial_{u}(\nabla_{} \cdot \nabla_{h} (\frac{1}{h}(f(u+iy)+h) - f(u+iy))) = -\partial_{v}(\nabla_{} \cdot \nabla_{h} (\frac{1}{h}(f(v+iy)+h) - f(v+iy))) $$
$$\partial_{u}( \partial_{\text{zzhh}} (\frac{1}{h}(f(u+iy)+h) - f(u+iy))) = \partial_{v}( \partial_{\text{zzhh}} (\frac{1}{h}(f(v+iy)+h) - f(v+iy)))$$
$$\partial_{u}( \partial_{zzhh} (\frac{1}{h}(f(u+iy)+h) - f(u+iy))) = -\partial_{\text{v}}( \partial_{\text{zzhh}} (\frac{1}{h}(f(v+iy)+h) - f(v+iy)))$$
Proof Begin by writing, for $z,w$ in a disc $B'$ with center $c$ $$f(z) - f(w) - f'(c)(z-w) = \int_{[z,w]} (f'(\xi)-f'(c))d\xi.$$ where $[z,w]$ is the straight line segment that joins $z$ and $w$. We can use this and the standard estimate${}^1$ to obtain $$\left|\frac{f(z)-f(w)}{z-w} - f'(c)\right| \leqslant |f'-f'(c)|_{[z,w]} \leqslant |f'-f'(c)|_{B'}$$ where the right hand side is the maximum of $f'-f'(c)$ over $B'$. Because $f'$ is continuous at $c$, it suffices we take $B'$ small enough to guarantee that $(1)$ holds.
$1$. By this I mean that $\left|\int_{\gamma} f dz\right|\leqslant L(\gamma)|f|_\gamma.$