Let $\boldsymbol{Y}$ be a matrix with dimension $2 \times n$, $\boldsymbol{X}_c$ and $\boldsymbol{X}_r$ are matrices $n \times p$, $\Sigma$ is a nonsingular and symmetric matrix $2 \times 2$, $\beta_c = (\beta_{c1}, \ldots, \beta_{cp})^T$ and $\beta_r = (\beta_{r1}, \ldots, \beta_{rp})^T$.
\begin{equation} \boldsymbol{u}_i =\{\boldsymbol{Y}_i - [\boldsymbol{X}_{ci}\beta_c, g^{-1}(\boldsymbol{X}_{ri}\beta_r)]^T\}^T \Sigma^{-1}\{\boldsymbol{Y}_i - [\boldsymbol{X}_{ci}\beta_c, g^{-1}(\boldsymbol{X}_{ri}\beta_r)]^T\}, i = 1, \ldots, n. \end{equation}
I'd like to calculate $\partial \boldsymbol{u}_i / \partial \boldsymbol{\beta}_r$. But, I cannot solve the differentiation with the $g^{-1}(\cdot)$ function.
I'll assume you know how to calculate the derivative of the inverse function, i.e. $$\eqalign{ h(z)=g^{-1}(z) \implies\frac{dh}{dz}= h'(z) \cr }$$ Define the vectors $$\eqalign{ e_k &\in {\mathbb R}^{n\times 1},\quad \varepsilon_k \in {\mathbb R}^{2\times 1}&{\rm (standard\,basis)} \cr z_k &= e_k^TX_r\beta_r,\quad p_k = h(z_k),\quad q_k = h'(z_k)&\in{\mathbb R}^{1}\cr w_k &= \pmatrix{e_k^TX_c\beta_c\\p_k} - \pmatrix{\varepsilon_1^TY_k\\\varepsilon_2^TY_k}&\in{\mathbb R}^{2\times 1} \cr &= \big(e_k^TX_c\beta_c-\varepsilon_1^TY_k\big)\varepsilon_1 + \big(p_k-\varepsilon_2^TY_k\big)\varepsilon_2 \cr dw_k &= \varepsilon_2\,dp_k = \varepsilon_2q_k\,dz_k \cr }$$ Write the $u_k$ vectors in terms of these new vectors. Then find the differential and gradient. $$\eqalign{ u_k &= w_k^T\Sigma^{-1}w_k = \Sigma^{-1}:w_kw_k^T \cr du_k &= 2\Sigma^{-1}w_k:dw_k \cr &= 2\Sigma^{-1}w_k:\varepsilon_2\,q_k\,dz_k \cr &= \Big(2\varepsilon_2^T\Sigma^{-1}w_k\Big)q_k\Big(e_k^TX_r\,d\beta_r\Big) \cr \frac{\partial u_k}{\partial\beta_r} &= \Big(2\varepsilon_2^T\Sigma^{-1}w_k\Big)q_k\Big(X_r^Te_k\Big) \cr &= \Big(2\varepsilon_2^T\Sigma^{-1}w_k\Big)\,X_r^TQe_k \cr }$$ where the matrix $\,Q={\rm Diag}(q_k)\in{\mathbb R}^{n\times n}$
In some intermediate steps, a colon was used to denote the trace/Frobenius product, i.e. $$\eqalign{A:B = {\rm Tr}(A^TB)}$$