= Solution
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 $r>0$,
$$
h_x=\frac{xy}{r},\qquad h_y=r+\frac{y^2}{r}-4.
$$
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 $\operatorname{diag}(1,2)$, and at $(0,-2)$ it is $\operatorname{diag}(-1,-2)$. At $(4,0)$ and $(-4,0)$ its <Hessian matrices> are respectively $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and $\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$. Therefore
$$
\boxed{\begin{array}{c|c|c}
(x,y)&g(x,y)&\text{type}\\\hline
(0,2)&-\sinh4&\text{strict local minimum}\\
(0,-2)&\sinh4&\text{strict local maximum}\\
(4,0),\ (-4,0)&0&\text{saddles}
\end{array}}
$$
The extrema are local: $g$ is unbounded above and below on the full <plane>.
For the <level curve> geometry, a level $g=c$ is $y(r-4)=\operatorname{arsinh}c$. 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 $-\sinh4$ 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.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-2-sinh-radial-contours.png]
{title=Level curves and all four stationary points of the radial hyperbolic-sine function}
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