= Solution
At a regular constrained stationary point, $\nabla g\ne0$ and the directional derivative of $f$ vanishes along the tangent to $g=0$. Therefore $\nabla f=\lambda\nabla g$ for a <Lagrange multiplier> $\lambda$. Solve this equation together with the constraint, then compare candidate values and any exceptional nonregular points. Proportionality of the two gradients gives
$$
\boxed{f_x(a,b)g_y(a,b)-f_y(a,b)g_x(a,b)=0.}
$$
If $\nabla g=0$ at an exceptional point, this determinant vanishes trivially, although the regular multiplier method need not identify every such point.
For the given ellipse, maximize $r^2=x^2+y^2$. The constraint is $r^2+xy=4$, and $xy\geq-r^2/2$, so $4\geq r^2/2$. Equality requires $x=-y$ and the constraint then gives $(x,y)=(2,-2)$ or $(-2,2)$. Thus
$$
\boxed{r_{\max}=2\sqrt2.}
$$
The constraint gradient vanishes only at the origin, which is not on this ellipse, so the regularity condition holds everywhere relevant.
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