Solve a PDE: $ x(y^2+z)p-y(x^2+z)q=(x^2-y^2)z$

since $$\dfrac{\dfrac{dx}{x}}{y^2+z}=\dfrac{\dfrac{dy}{y}}{-x^2-z}=\dfrac{dz}{x^2-y^2}$$ $$\Longrightarrow \dfrac{\dfrac{dx}{x}+\dfrac{dy}{y}}{y^2-x^2}=\dfrac{dz}{x^2-y^2}$$ $$\Longrightarrow \dfrac{dx}{x}+\dfrac{dy}{y}=-dz$$ $$\Longrightarrow z=-C\ln|xy|$$