【解】先计算偏导数。对 $z=\dfrac{x^2}{3y}+\arcsin(xy)$，有 $$\dfrac{\partial z}{\partial x} = \dfrac{2x}{3y} + \dfrac{y}{\sqrt{1-x^2y^2}},\quad \dfrac{\partial z}{\partial y} = -\dfrac{x^2}{3y^2} + \dfrac{x}{\sqrt{1-x^2y^2}}.$$ 则 $$xy\dfrac{\partial z}{\partial x} = xy\left(\dfrac{2x}{3y} + \dfrac{y}{\sqrt{1-x^2y^2}}\right) = \dfrac{2x^2}{3} + \dfrac{xy^2}{\sqrt{1-x^2y^2}},$$ $$y^2\dfrac{\partial z}{\partial y} = y^2\left(-\dfrac{x^2}{3y^2} + \dfrac{x}{\sqrt{1-x^2y^2}}\right) = -\dfrac{x^2}{3} + \dfrac{xy^2}{\sqrt{1-x^2y^2}}.$$ 相减得 $$xy\dfrac{\partial z}{\partial x} - y^2\dfrac{\partial z}{\partial y} = \left(\dfrac{2x^2}{3} + \dfrac{xy^2}{\sqrt{1-x^2y^2}}\right) - \left(-\dfrac{x^2}{3} + \dfrac{xy^2}{\sqrt{1-x^2y^2}}\right) = \dfrac{2x^2}{3} + \dfrac{x^2}{3} = x^2.$$