Understanding proof of analyticity of $\overline{f\left (\overline z\right )}$

I'm reading Marsden and Hoffman's book of Complex Analysis, and I'm struggling to understand the following example

Let $A$ be an open subset of $\mathbb{C}$ and $A^*=\{z|\overline z\in A\}$. Suppose $f$ is analytic on $A$, and define a function $g$ on $A^*$ by $g(z)=\overline{f\left (\overline z\right )}$. Show that $g$ is analytic on $A^*$.

The proof provided by the book is

If $f(z)=u(x,y)+iv(x,y)$, then $g(z)=\overline{f\left (\overline z\right )}=u(x,-y)-iv(x,-y)$.
We check the Cauchy-Riemann equations for $g$ as follows:$$\dfrac{\partial}{\partial x}(\operatorname{Re}g)=\dfrac{\partial}{\partial x}u(x,-y)=\left .\dfrac{\partial u}{\partial x}\right |_{(x,-y)}=\left .\dfrac{\partial v}{\partial y}\right |_{(x,-y)}=\dfrac{\partial}{\partial y}[-v(x,-y)]=\dfrac{\partial}{\partial y}(\operatorname{Im}g).$$

I'm failing to understand why $\left .\dfrac{\partial v}{\partial y}\right |_{(x,-y)}=\dfrac{\partial}{\partial y}[-v(x,-y)]$.
When I use the chain rule with $h(x,y)=(x,-y)$ I get $\dfrac{\partial}{\partial y}[v(x,-y)]=\dfrac{\partial}{\partial y}[v(h(x,y))]=\left .\dfrac{\partial (v\circ h)}{\partial y}\right |_{(x,y)}$ and\begin{align*}\begin{pmatrix}\dfrac{\partial}{\partial x}(v\circ h)&\dfrac{\partial}{\partial y}(v\circ h)\end{pmatrix} & =D(v\circ h) \\ & =D(v)D(h) \\ & =\begin{pmatrix}\dfrac{\partial}{\partial x}v&\dfrac{\partial}{\partial y}v\end{pmatrix}\begin{pmatrix}\dfrac{\partial}{\partial x}h_1&\dfrac{\partial}{\partial y}h_1\\\dfrac{\partial}{\partial x}h_2&\dfrac{\partial}{\partial y}h_2\end{pmatrix} \\ & =\begin{pmatrix}\dfrac{\partial}{\partial x}v&\dfrac{\partial}{\partial y}v\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix} \\ & =\begin{pmatrix}\dfrac{\partial}{\partial x}v&-\dfrac{\partial}{\partial y}v\end{pmatrix}. \end{align*}Hence $\dfrac{\partial}{\partial y}[v(x,-y)]=\left .\dfrac{\partial (v\circ h)}{\partial y}\right |_{(x,y)}=-\left .\dfrac{\partial v}{\partial y}\right |_{(x,y)}$.
So $\left .\dfrac{\partial v}{\partial y}\right |_{(x,y)}=\dfrac{\partial}{\partial y}[-v(x,-y)]=\left .\dfrac{\partial v}{\partial y}\right |_{(x,-y)}$, but the last one doesn't feel right.

Can anyone help me to prove that $\left .\dfrac{\partial v}{\partial y}\right |_{(x,y)}=\dfrac{\partial}{\partial y}[-v(x,-y)]$ or at least point out my mistake in the chain rule?

While doing the chain rule, you forgot to evaluate the matrix at each point. Remember that the chain rule says $D(f\circ g)(a)=D(f)(g(a))D(g)(a)$, not just $D(f\circ g)=D(f)D(g)$.