= Solution
Maximizing the squared distance avoids differentiating a square root. The <Lagrange multiplier> equations for $f=x^2+y^2$ are
$$
2x=\lambda(2x+y),\qquad2y=\lambda(x+2y),\qquad x^2+y^2+xy=4.
$$
Adding and subtracting the first two gives $(2-3\lambda)(x+y)=0$ and $(2-\lambda)(x-y)=0$. The feasible <stationary points> therefore lie on $x=y$ or $x=-y$. The first case gives $(x,y)=(\pm2/\sqrt3,\pm2/\sqrt3)$ and squared distance $8/3$. The second gives $(2,-2)$ and $(-2,2)$, with squared distance $8$.
The <ellipse> is compact and its constraint <gradient> does not vanish on the curve, so an extremum exists among these points. A direct global check is $4=f+xy\geq f/2$, using $xy\geq-f/2$. Equality occurs when $x=-y$. Consequently
$$
\boxed{d_{\max}=2\sqrt2,\qquad\text{at }(2,-2),\ (-2,2).}
$$
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