$
\def\bbR#1{{\mathbb R}^{#1}}
\def\d{\lambda}\def\g{\gamma}
\def\u#1{\underline{#1}}
\def\e{\varepsilon}\def\o{{\tt1}}\def\p{\partial}
\def\E{E_{\o\o}}
\def\M{M_{\o\o}}
\def\W{W_{\o\o}}
\def\V{V_{\o\o}}
\def\L{\left}\def\R{\right}
\def\LR#1{\L(#1\R)}\def\BR#1{\Big(#1\Big)}
\def\vec#1{\operatorname{vec}\LR{#1}}
\def\vech#1{\operatorname{vech}\LR{#1}}
\def\adj#1{\operatorname{adj}\LR{#1}}
\def\trace#1{\operatorname{Tr}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\c#1{\color{red}{#1}}
\def\mc#1{\left[\begin{array}{c}#1\end{array}\right]}
\def\m#1{\left[\begin{array}{r|rrr}#1\end{array}\right]}
$Let $E_{ij}\in\bbR{N\times N}$ denote a standard basis matrix with the $(i,j)^{th}$ element equal to $\o$ and all others equal to $0$.
Similarly use $\e_k\in\bbR{N^2\times\o}$ to denote a standard basis vector whose $k^{th}$ element equals $\o$.
Let's also use a colon to denote the Frobenius product, i.e.
$$\eqalign{
A:B &= \sum_{i=1}^n\sum_{j=1}^m A_{ij}B_{ij} \;=\; \trace{A^TB} \\
A:A &= \big\|A\big\|^2_F \\
}$$
This is also called the double-dot or double contraction product.
When applied to vectors $(n=N^2,\,m=\o)$ it reduces to the standard dot product.
When applied to square matrices $(n=N,\,m=N)$ the trace definition is convenient.
The function in question can be rewritten in a form which is easily differentiated.
$$\eqalign{
g &= \E:V \\
&= \vec{\E}:\vec{V} \\
&= {\e_{\o}}:v \\
dg &= {\e_{\o}}:dv \\
\grad{g}{v} &= {\e_{\o}} \\
}$$
In the symmetric case, let
$$\eqalign{
w &= \vech{V}, \quad
v &= \vec{V} &= Dw \\
}$$
where $D$ is the Duplication matrix
Then by a similar calculation
$$\eqalign{
g &= {\e_{\o}}:v \\
&= {\e_{\o}}:Dw \\
&= D^T{\e_{\o}}:w \\
dg &= D^T{\e_{\o}}:dw \\
\grad{g}{w} &= D^T{\e_{\o}} \qquad\qquad\qquad \\
}$$
which is just the first basis vector
for the half-vec space $\bbR{N(N+1)/2}$
Update #1
The comments pointed out that underlined quantities explicitly exclude the $\V$ element, e.g.
$$\u{v}=\u{\rm vec}(V)$$
For these underlined vec/vech operators, both of the derivatives are zero vectors of dimensions $\,\bbR{N^2-\o}\,$ and $\,\bbR{N(N+\o)/2-\o}\,$ respectively.
Update #2
Based on the comments, the intent of the question is to consider $\V$ as an implicit function of the remaining elements of $V$ based on the constraint $\,\det(V)=\o.\,$
The first step is to reverse the vectorizations and reconstitute the matrix. Since $(\o,\o)$ element is not part of the vectors, the respective unvec operators will set it to zero in the reconstituted matrix, i.e.
$$\eqalign{
&M = \u{\rm unvec}(\u{v}) = \u{\rm unvech}(\u{w}) \\
&\M = 0 \\
}$$
Calculate the inverse and determinant of this reconstituted matrix.
$$\eqalign{
W &= M^{-1} \\
\d &= \det(M) \qiq \g=\det(W)=\d^{-1} \\
}$$
The remaining task is to calculate $\M$ such that the constraint $\d=\o$ is satisfied.
The Jacobi formula tells us that
$$\eqalign{
\d &= \det(M) \qiq d\d = \d W^T:dM \\
}$$
If the change $dM$ is restricted to its $(\o,\o)$ element then we are left with the scalar equation
$$\eqalign{
d\d &= \d \W\;d\M \\
}$$
The increment $d\d=(\o-\d)\,$ yields $\,(\d+d\d)=\o,\,$
which satisfies the constraint.
Therefore incrementing the $\M$ element by
$$\eqalign{
d\M &= \frac{\o-\d}{\d\W} \qiq V = M + \E\,d\M \\
}$$
will satisfy the constraint, assuming that neither
$\W$ nor $\d$ is equal to zero. For the proposed
${V},\,$ applying the Matrix Determinant Lemma will verify that indeed $\,\det({V})=\o$.
Now we can write an explicit expression for the $g$-function
$$\eqalign{
g = \V
= \LR{\frac{\o-\d}{\d\W}}
= \LR{\frac{\det(W)-\o}{\E:W}}
\\
}$$
which we can differentiate
$$\eqalign{
dg &= \LR{\frac{d\det(W)}{\E:W}}
- \LR{\frac{\det(W)-\o}{(\E:W)^2}}\BR{\E:dW} \\
&= \LR{\frac{\g M:dW}{\E:W}}
- \LR{\frac{\g-\o}{(\E:W)^2}}\BR{\E:dW} \\
&= \LR{\frac{(\E:W)\g M-(\g-\o)\E}{(\E:W)^2}}:dW \\
&= \LR{\frac{(\g-\o)\E-\g\W M}{\W^2}}:W\,dM\,W \\
&= W\LR{\frac{(\g-\o)\E-\g\W M}{\W^2}}W:dM \\
&= \LR{\frac{(\g-\o)\LR{W\E W}-\g\W W}{\W^2}}:dM \\
&=\u{\rm vec}\LR{\frac{(\g-\o)\LR{W\E W}-\g\W W}{\W^2}}:d\u{v} \\
\grad{g}{\u{v}}
&=\u{\rm vec}\LR{\frac{(\g-\o)\LR{W\E W}-\g\W W}{\W^2}} \\
}$$
The Duplication matrix immediately yields the other derivative
$$\eqalign{
\grad{g}{\u{w}} &= \u{D}^T \LR{\grad{g}{\u{v}}} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad \\
\\
}$$
NB: The following matrix removes the first element of
an $n$-vector
$$\eqalign{
R(n) = \m{ 0_{n-1} & I_{n-1}},
\quad {\rm e.g.}\;\;
R(4) = \m{
0 & \o & 0 & 0 \\
0 & 0 & \o & 0 \\
0 & 0 & 0 & \o \\
} \in \bbR{3\times 4}
\\
}$$
For typing convenience, let $R=R\BR{N^2}$
and $Q=R\BR{N(N+\o)/2}$
This allows all of the underlined symbols to be
expressed in standard matrix notation
$$\eqalign{
\u{w} &= \u{\rm vech}(V) &= Q\;\vech{V} &= Qw \\
\u{v} &= \u{\rm vec}(V) &= R\;\vec{V} &= Rv \\
\u{D} &= RDQ^T \\
}$$
$$\eqalign{
M &= \u{\rm unvech}(\u{w}) &= {\rm unvech}(Q^T\u{w})\quad \\
&= \u{\rm unvec}(\u{v}) &= {\rm unvec}(R^T\u{v}) \\
}$$
Update #3
Exploiting the block structure of the matrix to evaluate the determinant
yields an alternate expression for the $g$-function.
$$\eqalign{
V &= \mc{\V & x^T \\ x&Y} \;=\; V^T \\
\det(V) &= \LR{\V-x^TY^{-1}x} \det(Y) \;&\doteq\; \large\o \\
\V &= x^TY^{-1}x + \det(Y^{-1}) \;&\doteq\; g \\
}$$
If you differentiate this expression you will find
$$\eqalign{
dg
&= 2Y^{-1}x:dx - Y^{-1}\LR{xx^T+\frac I{\det Y}}Y^{-1}:dY \\\\
&= \mc{
2Y^{-1}x \\
-D^T\LR{Y\otimes Y}^{-1}\vec{xx^T+\frac I{\det Y}} \\
}:d\u{v} \\\\
\grad{g}{\u{v}}
&= \mc{
2Y^{-1}x \\
-D^T\LR{Y\otimes Y}^{-1}\vec{xx^T+\frac I{\det Y}} \\
} \\\\
}$$
Note that
$$\eqalign{
{\rm vech}(V)
= \mc{\V \\ x \\ {\rm vech}(Y)}
= \mc{\V \\ x \\ y}
,\qquad
\u{v}
= \u{\rm vech}(V)
= \mc{x \\ y}
\\
}$$
Best Answer
Let $F(x):=W\mathrm{diag}(f(Ax+b))$, then we can simply rewrite it as
$$F(x)=W\sum_{j=1}^Ke_je_j^Tf(e_j^TAx+b_j)$$
where $\{e_i\}$ is the natural basis for $\mathbb{R}^K$ and $b=[b_i]$. This is a simple rewriting of
The most convenient representation (at least to me) of the derivative with respect to $x$ is then
$$\dfrac{\partial F(x)}{\partial x_i}=W\sum_{j=1}^Ke_je_j^Ta_{ji}f'(e_j^TAx+b_j)$$
where $A=[a_{ij}]$, $f'$ is the derivative of $f$, and where we have applied the usual derivative rules of derivation.
This can be rewritten in terms of the initial "diag" operator as
$$\dfrac{\partial F(x)}{\partial x_i}=WD_i\mathrm{diag}(f'(Ax+b))$$
where $D_i:=\mathrm{diag}(a_{1i},\ldots,a_{Ki})$.