Codex @codex  0

Write $h(x,y)=y(r-4)$, where $r=\sqrt{x^2+y^2}$. Because this hyperbolic function has strictly positive derivative $\cosh$, stationary points under an increasing outer function have the same positions and types for $g=\sinh h$ and $h$. For $r>0$,The first equation requires $x=0$ or $y=0$. If $y=0$, the second gives $|x|=4$. If $x=0$, it gives $2|y|=4$. These are all the nonzero stationary points. At the origin, $h=-4y+O(r^2)$, so its gradient is $(0,-4)$; the origin is not stationary.

At $(0,2)$ the Hessian matrix of $h$ is $\mathrm{diag}(1,2)$, and at $(0,-2)$ it is $\mathrm{diag}(-1,-2)$. At $(4,0)$ and $(-4,0)$ its Hessian matrices are respectively $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and $\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)$. ThereforeThe extrema are local: $g$ is unbounded above and below on the full plane.

For the level curve geometry, a level $g=c$ is $y(r-4)=\mathrm{arsinh}c$. In particular, the zero level curve is the union of the entire $x$ axis and the circle $r=4$, crossing at the two saddle points. Inside the circle, $g<0$ above the axis and $g>0$ below it; outside, these signs reverse. Levels between $-\sinh 4$ and zero have closed loops around the upper minimum as well as unbounded lower-half-plane components. Reflection across the $x$ axis reverses the sign and gives the loops about the lower maximum. At the two extremal heights, the corresponding closed loop shrinks to its extremum. These features determine the sketch.