 Solution

Differentiate $v_t=y{u}_x-x{u}_y$ with respect to $t$, remembering that its coefficients also depend on $t$. The product rule and multivariable chain rule giveFor a twice continuously differentiable $u$, equality of mixed partial derivatives gives $u_{yx}=u_{xy}$. HenceThe first-order terms come from differentiating $x$ and $y$ themselves; treating $y{\partial }_x-x{\partial }_y$ as an operator with constant coefficients would miss them.