Solution: We recall that if a function $f:\mathbb{R} ^m \to \mathbb{R} ^n$ is given by \[ f(\mathbf{x} ) = \left( f_1(\mathbf{x} ),f_2(\mathbf{x} ),\ldots ,f_n(\mathbf{x} ) \right), \] where $\mathbf{x} = \left( x_1,x_2,\ldots , x_m \right) $, then the derivative at $\mathbf{x} $ will be \[ df_{\left( \mathbf{x} \right) } \coloneqq \begin{pmatrix} \frac{\partial f_1}{\partial x_1}(\mathbf{x}) & \frac{\partial f_1}{\partial x_2}(\mathbf{x}) & \cdots & \frac{\partial f_1}{\partial x_m}(\mathbf{x}) \\[2ex] \frac{\partial f_2}{\partial x_1}(\mathbf{x}) & \frac{\partial f_2}{\partial x_2}(\mathbf{x}) & \cdots & \frac{\partial f_2}{\partial x_m}(\mathbf{x}) \\[2ex] \vdots & \vdots & \cdots & \vdots \\ \frac{\partial f_n}{\partial x_1}(\mathbf{x}) & \frac{\partial f_n}{\partial x_2}(\mathbf{x}) & \cdots & \frac{\partial f_n}{\partial x_m}(\mathbf{x}) \\[2ex] \end{pmatrix}_{n\times m}. \]

Here the function is \[ f(x,y) = \left( x^2 , y^2 + \sin x \right) = \left( f_1(x,y),f_2(x,y) \right). \] Therefore, \begin{align*} df_{(2,3)} & = \begin{pmatrix} \frac{\partial f_1}{\partial x}(2,3) & \frac{\partial f_1}{\partial y}(2,3) \\[2ex] \frac{\partial f_2}{\partial x}(2,3) & \frac{\partial f_2}{\partial y}(2,3) \\ \end{pmatrix} \\[1ex] & = \left.\begin{pmatrix} 2x & 0 \\ \cos x & 2y \\ \end{pmatrix}\right|_{(2,3)} \\[1ex] & = \begin{pmatrix} 4 & 0 \\ \cos 2 & 6 \\ \end{pmatrix}. \end{align*}