Solution: Recall that a general equation of a circular cone of height $h$ and base radius $r$ oriented along the $z$-axis with vertex pointing up is given by \begin{equation}\label{eq:30May2025-1} x^2 + y^2 = (z- z_0)^2 \tan^2 \theta, \end{equation} where $\tan \theta = \frac{r}{h}$. To determine the corresponding partial differential equation, we need to eliminate $z_0$ and $\theta $. Take the partial derivative of \eqref{eq:30May2025-1} with respect to $x$ and $y$, \begin{align} 2x = 2(z - z_0) \frac{\partial z}{\partial x} \tan ^2 \theta & \implies x = (z - z_0) \frac{\partial z}{\partial x} \tan ^2 \theta \label{eq:30May2025-2} \\ 2y = 2(z - z_0) \frac{\partial z}{\partial y} \tan ^2 \theta & \implies y = (z - z_0) \frac{\partial z}{\partial y} \tan ^2 \theta. \label{eq:30May2025-3} \end{align}

Multiplying \eqref{eq:30May2025-2} by $y$ and \eqref{eq:30May2025-3} by x gives, \begin{align*} xy & = y(z-z_0)\frac{\partial z}{\partial x} \tan ^2 \theta \quad \text{ and } \\ xy & = x(z-z_0)\frac{\partial z}{\partial y} \tan ^2 \theta. \end{align*} This implies, \begin{align*} & y(z-z_0)\frac{\partial z}{\partial x} \tan ^2 \theta = x(z-z_0)\frac{\partial z}{\partial y} \tan ^2 \theta \\ \implies & y \frac{\partial z}{\partial x} = x \frac{\partial z}{\partial y} \implies y \frac{\partial z}{\partial x} - x \frac{\partial z}{\partial y} = 0. \end{align*} Thus, the required partial differential equation will be \[ \textcolor{blue}{ \boxed{ y \frac{\partial z}{\partial x} - x \frac{\partial z}{\partial y} = 0 } } \]