1. The problem is to solve the partial differential equation using Lagrange's method: $$yzp - xzq = xy$$ where $p = \frac{\partial z}{\partial x}$ and $q = \frac{\partial z}{\partial y}$. 2. Write the equation in the form: $$\frac{dx}{yz} = \frac{dy}{-xz} = \frac{dz}{xy}$$ 3. From the first two ratios: $$\frac{dx}{yz} = \frac{dy}{-xz} \implies -x z dx = y z dy \implies -x dx = y dy$$ (dividing both sides by $z$) 4. Simplify and integrate: $$-x dx = y dy \implies -\int x dx = \int y dy \implies -\frac{x^2}{2} = \frac{y^2}{2} + C_1$$ 5. Multiply both sides by 2: $$-x^2 = y^2 + 2C_1 \implies x^2 + y^2 = C$$ where $C = -2C_1$ is a constant. 6. From the second and third ratios: $$\frac{dy}{-xz} = \frac{dz}{xy} \implies y x dz = -x z dy \implies y dz = -z dy$$ (dividing both sides by $x$) 7. Rearrange: $$\frac{dz}{z} = -\frac{dy}{y}$$ 8. Integrate both sides: $$\int \frac{dz}{z} = -\int \frac{dy}{y} \implies \ln|z| = -\ln|y| + C_2$$ 9. Exponentiate both sides: $$z = \frac{C_3}{y}$$ where $C_3 = e^{C_2}$ is a constant. 10. The two independent integrals are: $$x^2 + y^2 = C_1$$ and $$z y = C_2$$ 11. The general solution is an arbitrary function of these integrals: $$F(x^2 + y^2, z y) = 0$$ Thus, the solution to the PDE is given implicitly by $$F(x^2 + y^2, z y) = 0$$ where $F$ is an arbitrary function.