Solution: We write $\mathbf{x} = (x_1, x_2, x_3, x_4)$. Let $\mu_i = \lambda x_i $. Differentiating the expression $f(\lambda \mathbf{x} ) = \lambda ^3 f(\mathbf{x} )$ with respect to $\lambda $ we get \begin{equation}\label{eq:27May2024-1} \sum_{i=1}^{4} \frac{\partial f}{\partial \mu_i} x_i = 3\lambda ^2 f(\mathbf{x} ) \end{equation} Similarly, differentiating with respect to $x_i$'s, we get \begin{align*} \lambda \frac{\partial f}{\partial \mu_i} = \lambda ^3 \frac{\partial f}{\partial x_i}, \quad i = 1,2,3,4. \\ \implies \frac{\partial f}{\partial \mu_i} = \lambda ^2 \frac{\partial f}{\partial x_i} ,\quad i = 1,2,3,4. \end{align*} Substituting the value of $\frac{\partial f}{\partial \mu_i} $ in \eqref{eq:27May2024-1}, we obtain \begin{align*} \sum_{i=1}^{4} \left( \lambda ^2 \frac{\partial f}{\partial x_i} \right) x_i = 3\lambda ^2 f(\mathbf{x} ) \implies \sum_{i=1}^{4} x_i \frac{\partial f}{\partial x_i}(\mathbf{x} ) = 3f(\mathbf{x} ). \end{align*}

Take $\mathbf{x} = \left( 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4} \right)$, we get \begin{align*} & \frac{\partial f}{\partial x_1} (p) + \frac{1}{2} \frac{\partial f}{\partial x_2} (p) + \frac{1}{3}\frac{\partial f}{\partial x_3}(p) + \frac{1}{4}\frac{\partial f}{\partial x_4} (p) = 3f(p) \\ \implies & 12\frac{\partial f}{\partial x_1} (p) + 6 \frac{\partial f}{\partial x_2} (p) + 4\frac{\partial f}{\partial x_3}(p) + 3\frac{\partial f}{\partial x_4} (p) = 36 f(p) = 216. \end{align*} Thus, \[ 12\frac{\partial f}{\partial x_1} (p) + 6\frac{\partial f}{\partial x_2} (p) + 4 \frac{\partial f}{\partial x_3} + 3 \frac{\partial f}{\partial x_4} = 216. \]