Solution: The given equation is \begin{equation}\label{eq:21Feb2025-1} p^2 - q^2 = \left( x^2 + y^2 \right) z, \end{equation} which can be re-written as \begin{equation}\label{eq:21Feb2025-2} \left( \frac{1}{\sqrt{z} } \frac{\partial z}{\partial x} \right) ^2 + \left( \frac{1}{\sqrt{z}}\frac{\partial z}{\partial y} \right) ^2 = x^2 + y^2. \end{equation} Let \[ \frac{1}{\sqrt{z}}\mathrm{d} z = \mathrm{d} Z \implies 2 \sqrt{z} = Z . \] Using the above in \eqref{eq:21Feb2025-2}, we get \[ \left( \frac{\partial Z}{\partial x} \right) ^2 + \left( \frac{\partial Z}{\partial y} \right) ^2 = x^2 + y^2, \] which we can write as \[ P^2 + Q^2 = x^2 + y^2 , \quad P = \frac{\partial Z}{\partial x} \text{ and } Q = \frac{\partial Z}{\partial y} . \]

Thus we get \begin{align*} P^2 - x^2 = y^2 - Q^2. \end{align*} This implies both the equation will be constant. Equating each side to an arbitrary constant $a^2$, we have \begin{align*} & P^2 - x^2 = a^2 \quad \text{ and } \quad y^2 - Q^2 = a^2 \\ \implies & P = \sqrt{x^2 + a^2} \quad \text{ and } \quad Q = \sqrt{y^2 - a^2} . \end{align*} Since \begin{align*} \mathrm{d} Z = P \mathrm{d} x + Q \mathrm{d} y & \implies \mathrm{d} Z = \sqrt{x^2 + a^2} \mathrm{d} x + \sqrt{y^2 - a^2} \mathrm{d} y \\ & \implies Z = \textcolor{blue}{\int \sqrt{x^2 + a^2}\ \mathrm{d} x} + \textcolor{magenta}{\int \sqrt{y^2 - a^2}\ \mathrm{d} y} \\ & \implies Z = \textcolor{blue}{\frac{1}{2} x \sqrt{x^2 + a^2} + \frac{a^2}{2} \ln \left\vert x + \sqrt{x^2 + a^2} \right\vert }\\ & \kern 1.7cm + \textcolor{magenta}{\frac{1}{2} y \sqrt{y^2 - a^2} - \frac{a^2}{2} \ln \left\vert x + \sqrt{y^2 - a^2} \right\vert}. \end{align*} Since $Z = 2\sqrt{z} $,the required integral will be \[ z = \frac{Z^2}{4}. \]