Let $\mathbf{I}_m$ be the $m$-dimensional identity matrix and $\mathbf{0}_m$ be the $m$-dimensional zero matrix. The matrix $D(\mathbb{x})$, where $\mathbb{x} = (x_1, \dots, x_n)^T$, is defined as: $$ D(\mathbb{x}) = \begin{pmatrix} x_1 \mathbf{I}_{m} & \mathbf{0}_m & \mathbf{0}_m & \dots & \mathbf{0}_m \\ \mathbf{0}_m & x_2 \mathbf{I}_{m} & \mathbf{0}_m & \dots & \mathbf{0}_m \\ \mathbf{0}_m & \mathbf{0}_m & x_3 \mathbf{I}_{m} & \dots & \mathbf{0}_m \\ \vdots & \vdots & \vdots & \dots & \vdots \\ \mathbf{0}_m & \mathbf{0}_m & \mathbf{0}_m & \dots & x_n \mathbf{I}_{m} \end{pmatrix} $$ That is, $D(\mathbf{x})$ is a diagonal matrix with entries from $\mathbf{x}$ on the diagonal (each entry is replicated $m$ times).
Now, given a vector $\mathbf{v}$ and a positive-definite matrix $\mathbf{P}$, we define a function on the domain $\mathbf{x} \geq 0$: $$ f(\mathbf{x}) = \mathbf{v}^T (\mathbf{P} + D(\mathbf{x}))^{-1} \mathbf{v} $$ Is there a way to express the gradient and Hessian of $f$?
First, let's define $$Q(x)=P + D(x) = P + \mathop{\textrm{diag}}(x) \otimes I_m$$ where $\mathop{\textrm{diag}}$ maps the vector $x$ to the corresponding diagonal matrix, and $\otimes$ denotes the Kronecker product. This makes it a bit simpler to see that $$\frac{\partial Q(x)}{\partial x_i} = \frac{\partial D(x)}{\partial x_i} = e_ie_i^T \otimes I_m = (e_i \otimes I_m) (e_i^T \otimes I_m)$$ where $e_i$ is unit vector with a $1$ in the $i$th element. Expressed without the Kronecker product, $e_ie_i^T\otimes I_m$ is the block matrix with $I_m$ in the $(i,i)$ block position and $0_m$ everywhere else; and $e_i \otimes I_m$ is the $i$th block column of that matrix.
Armed with this information, we proceed per the aforementioned Matrix Cookbook. From equation 40, we have $$\begin{aligned} \frac{\partial Q(x)^{-1}}{\partial x_i} &= - Q(x)^{-1} (e_ie_i^T \otimes I_m) Q(x)^{-1} \\ &= - Q(x)^{-1} (e_i \otimes I_m) (e_i^T \otimes I_m) Q(x)^{-1} \end{aligned}$$ We can obtain a very convenient interpretation of this quantity if we apply a corresponding block structure to $Q(x)^{-1}$. Let's call this quantity $R$, so that $$R \triangleq Q(x)^{-1} = \begin{bmatrix} R_{11} & R_{12} & \dots & R_{1n} \\ R_{21} & R_{22} & \dots & R_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ R_{n1} & R_{n2} & \dots & R_{nn} \end{bmatrix}$$ With this structure in place, we see that right-multiplying $Q(x)^{-1}$ by $e_i \otimes I_m$ obtains the $i$th block column. Let's denote that column by $R_{i:}$.
This leads us to this first derivative: $$\begin{aligned} \frac{\partial f(x)}{\partial x_i} &= - v^TQ(x)^{-1} (e_ie_i^T \otimes I_m) Q(x)^{-1}v \\ &= - v^TQ(x)^{-1} (e_i \otimes I_m) (e_i^T \otimes I_m) Q(x)^{-1}v \\ &= - v^T R_{i:} R_{i:}^T v = \|R_{i:}^T v\|_2^2\end{aligned}$$ The second partial derivatives are $$\begin{aligned} \frac{\partial^2 f(x)}{\partial x_i x_j} &= 2 v^TQ(x)^{-1} (e_ie_i^T \otimes I_m)Q(x)^{-1} (e_je_j^T \otimes I_m) Q(x)^{-1}v \\ &= 2 v^TQ(x)^{-1} (e_i \times I_m)(e_i^T \otimes I_m)Q(x)^{-1} (e_j\otimes I_m)(e_j^T \otimes I_m) Q(x)^{-1}v \\ &= 2 v^T R_{i:} R_{ij} R_{j:}^T v \end{aligned}$$where $R_{ij}$ is the $(i,j)$ block of $Q(x)^{-1}$.