Solution: The given equation is \begin{equation}\label{eq:28Mar2025-1} x\left( y^2 +z\right)p - y\left( x^2 + z \right)q = \left( x^2 - y^2 \right) z . \end{equation} The corresponding Lagrange's auxiliary equations are \begin{equation}\label{eq:28Mar2025-2} \frac{\ \mathrm{d} x}{x \left( y^2 + z \right) } = \frac{\ \mathrm{d} y}{-y \left( x^2 + z \right) } = \frac{\ \mathrm{d} z}{z \left( x^2 - y^2 \right) }. \end{equation} Choosing $\frac{1}{x}, \frac{1}{y}, \frac{1}{z} $ as multipliers, each fraction of \eqref{eq:28Mar2025-1} equals to \begin{align*} \frac{\frac{1}{x} \ \mathrm{d}x + \frac{1}{y}\ \mathrm{d} y + \frac{1}{z}\ \mathrm{d} z}{y^2 + z - \left( x^2 + z \right) + x^2 - y^2} = \frac{\frac{1}{x}\ \mathrm{d} x + \frac{1}{y}\ \mathrm{d} y + \frac{1}{z}\ \mathrm{d} z}{0}, \end{align*} which implies \begin{align} \frac{1}{x} \ \mathrm{d} x + \frac{1}{y}\ \mathrm{d} y + \frac{1}{z}\ \mathrm{d} z = 0 & \implies \ln (x) + \ln (y) + \ln (z) = \ln (c_1) \notag\\ & \implies \ln (xyz) = \ln (c_1)\notag \\ & \implies xyz = c_1. \label{eq:28Mar2025-3} \end{align} Similarly, choosing $x,y,-1$ as multipliers, each fraction of \eqref{eq:28Mar2025-1} equals to \begin{align*} \frac{x \ \mathrm{d} x + y \ \mathrm{d} y - \ \mathrm{d} z}{x^2 (y^2 + z) - y^2 (x^2 + z) - z(x^2 - y^2)} = \frac{x \ \mathrm{d} x + y \ \mathrm{d} y + \ \mathrm{d} }{0}, \end{align*} which implies \begin{align} \label{eq:28Mar2025-4} x \ \mathrm{d} x + y \ \mathrm{d} y - \ \mathrm{d} z = 0 & \implies x^2 + y^2 - 2z = c_2. \end{align}

Now from the given straight line $x+ y= 0, z = 1$, we can write, \[ x = t, \quad y = -t,, \text{ and } \quad z = 1. \] Thus, the solutions can be written as \begin{gather*} xyz = c_1 \implies -t^2 = c_1 \\ x^2 + y^2 - 2z = c_2 \implies 2t^2 - 2 = c_2. \end{gather*} Solving for $t$, we get \begin{align*} 2(-c_1) - 2 = c_2 \implies 2c_1 + c_2 + 2 = 0. \end{align*} Finally, substituting $c_1$ and $c_2$ from \eqref{eq:28Mar2025-3} and \eqref{eq:28Mar2025-4}, the integral surface is \[ \textcolor{blue}{ \boxed{ 2xyz + x^2 + y^2 - 2z + 2 = 0. } } \]