= Solution
Let $S_{s,t}(y)$ be the position at time $t$ of the <characteristic curve> satisfying $X(s)=y$ and $\dot X=u(t,X)$. Smooth uniqueness implies $S_{t,s}=S_{s,t}^{-1}$. Differentiating in the initial point gives the <variational equation>
$$
\partial_t D_yS_{0,t}
=Du(t,S_{0,t})D_yS_{0,t},\qquad D_yS_{0,0}=I.
$$
The <Liouville formula for a fundamental matrix>, or the derivative formula for a <determinant>, therefore gives
$$
\partial_t\det D_yS_{0,t}
=(\operatorname{div}u)(t,S_{0,t})\det D_yS_{0,t}.
$$
Because $u=(-\phi_{x_2},\phi_{x_1})$, the mixed second derivatives cancel and $\operatorname{div}u=0$. The initial <Jacobian determinant> is one, so
$$
\boxed{\det D_yS_{0,t}(y)=1.}
$$
Thus this <Hamiltonian flow> preserves planar <Lebesgue measure>. This calculation assumes smoothness as in the part; a rough-flow conclusion later must be justified by approximation rather than by differentiating a merely continuous velocity.
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