= Solution
Differentiate $v_t=yu_x-xu_y$ with respect to $t$, remembering that its coefficients also depend on $t$. The <product rule> and <multivariable chain rule> give
$$
\begin{aligned}
v_{tt}&=y_tu_x+y(u_{xx}x_t+u_{xy}y_t)-x_tu_y-x(u_{yx}x_t+u_{yy}y_t)\\
&=-xu_x+y(yu_{xx}-xu_{xy})-yu_y-x(yu_{yx}-xu_{yy}).
\end{aligned}
$$
For a twice continuously differentiable $u$, equality of <mixed partial derivatives> gives $u_{yx}=u_{xy}$. Hence
$$
\boxed{v_{tt}=y^2u_{xx}-2xyu_{xy}+x^2u_{yy}-xu_x-yu_y.}
$$
The 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.
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