= Solution
Using the coordinate formula for the <Lie bracket of vector fields>, the two components are
$$
[X,Y]^x=X(0)-Y(y)=-\frac{x^2}{2},\qquad
[X,Y]^y=X(x^2/2)-Y(0)=xy.
$$
Thus $\boxed{[X,Y]=-(x^2/2)\partial_x+xy\partial_y}$, which is not the zero <vector field>.
The <integral curves of a vector field> $X$ solve $\dot x=y$, $\dot y=0$, whereas those of $Y$ solve $\dot x=0$, $\dot y=x^2/2$. Their <local flows> are
$$
\boxed{\phi_t(x,y)=(x+ty,y),\qquad \psi_s(x,y)=(x,y+sx^2/2).}
$$
Both formulas exist for all real parameters, so both are <complete vector fields>. Direct composition gives
$$
\phi_t\psi_s(x,y)=(x+ty+tsx^2/2,\ y+sx^2/2),\qquad
\psi_s\phi_t(x,y)=(x+ty,\ y+s(x+ty)^2/2).
$$
At $(1,0)$ their first coordinates differ by $ts/2$. Hence these <local flows> do not commute when $s,t$ are nonzero.
For the <Lie bracket of vector fields>, start at $(2,0)$. An <integral curve of a vector field> $[X,Y]$ is
$$
(x(t),y(t))=\left(\frac{2}{1+t},0\right),\qquad -1<t<\infty.
$$
It solves $\dot x=-x^2/2$, $\dot y=xy$, but $x(t)$ diverges as $t\downarrow-1$. It cannot extend through that finite time as a curve in $\mathbb R^2$. \b[Complete vector fields are therefore not closed under the Lie bracket.]
For the sum $X+Y$, its <integral curves of a vector field> obey $\dot x=y$, $\dot y=x^2/2$. Substitution of $x=(1+\alpha t)^{-2}$ and $y=\beta(1+\alpha t)^{-3}$ requires
$$
\beta=-2\alpha,\qquad -3\alpha\beta=\frac12,
\qquad \alpha^2=\frac1{12}.
$$
Choosing $\alpha=-1/(2\sqrt3)$ gives the <integral curve of a vector field>
$$
\gamma(t)=\left(\left(1-\frac{t}{2\sqrt3}\right)^{-2},
\frac1{\sqrt3}\left(1-\frac{t}{2\sqrt3}\right)^{-3}\right),\qquad t<2\sqrt3.
$$
It starts at $(1,1/\sqrt3)$ and escapes to infinity at the finite time $2\sqrt3$. \b[Complete vector fields are not closed under addition either.] Both failures use the same pair of <complete vector fields>.
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