$\def\e{{\bf e}}
\def\D{\mathrm{diag}(\x)}
\def\x{{\bf x}}
\def\A{{\bf A}}
\def\M{{\bf M}}
\def\Mi{{\bf M}^{-1}}
\def\P{{\bf P}_i}
\def\id{\mathbb{I}}
\def\tr{\mathrm{Tr}\,}$
Let $\M = \A + \D$.
Note that $d(\M\Mi) = d \id = {\bf 0}$.
Thus, $d\Mi = -\Mi d \M \Mi$, and so
$$\begin{eqnarray*}
d \tr \Mi &=& d \sum \e_i^T \Mi \e_i \\
&=& \sum \e_i^T d \Mi \e_i \\
&=& -\sum \e_i^T \Mi d \M \Mi \e_i.
\end{eqnarray*}$$
If $\A$ is a function of $\x$, this is about as far as we'll go,
$$\begin{eqnarray*}
\frac{\partial}{\partial x_i} \tr \Mi
&=& -\sum_j \e_j^T \Mi \frac{\partial \M}{\partial x_i} \Mi \e_j \\
&=& -\tr \left((\Mi)^2 \frac{\partial \M}{\partial x_i}\right).
\end{eqnarray*}$$
If $\A$ is not a function of $\x$ we have that $\P = \frac{\partial \M}{\partial x_i}$ is a projection operator.
(All components of $\P$ are zero except the $ii$th component, which is $1$.)
In that case we find
$$\begin{eqnarray*}
\frac{\partial}{\partial x_i} \tr \Mi
&=& -\sum_j \e_j^T \Mi \P \Mi \e_j \\
&=& -\tr \left((\Mi)^2 \P\right) \\
&=& -(\Mi)^2_{ii} \\
&=& -\sum_j \Mi_{ij}\Mi_{ji}.
\end{eqnarray*}$$
In terms of $\A$, $\x$, and $\e_i$,
$$\begin{eqnarray*}
\frac{\partial}{\partial x_i} f(\x)
&=& -\sum_j \e_j^T (\A + \D)^{-1} \frac{\partial \D}{\partial x_i} (\A + \D)^{-1} \e_j \\
&=& -\sum_j \e_j^T (\A + \D)^{-1} \P (\A + \D)^{-1} \e_j.
\end{eqnarray*}$$
As noted by @passerby51 we could find an expression for $(\A + \D)^{-1}$ in terms of $\A^{-1}$ and $\D^{-1}$, but we will stop here.
Addendum: We made no assumptions about the basis $\e_i$.
If $\e_i$ is the natural basis the above implies
$$\begin{eqnarray*}
\frac{\partial}{\partial x_i} f(\x)
&=& -\e_i^T (\A + \D)^{-2} \e_i,
\end{eqnarray*}$$
which agrees with @passerby51's result.
(Note that in this case $\P = \e_i \e_i^T$.)
By the Woodbury formula,
$$\begin{eqnarray*}
(\A+\D)^{-1} &=& \A^{-1} - \A^{-1}(\A^{-1} + \D^{-1})^{-1} \A^{-1}.
\end{eqnarray*}$$
This may or may not be useful depending on the form of $\A$.
