= Solution
Put $s=x^2+y^2$. The <gradient> components are
$$
f_x=4x(s-1),\qquad f_y=y\left[4(s-1)+\frac7{3+y^2}\right].
$$
A <stationary point> requires both components to vanish. If $x\ne0$, then $s=1$; the second component subsequently forces $y=0$, giving $(\pm1,0)$. If $x=0$, either $y=0$ or, writing $u=y^2$,
$$
4(u-1)(3+u)+7=4u^2+8u-5=(2u-1)(2u+5)=0.
$$
Only $u=1/2$ is nonnegative. Thus there are five <stationary points>, with values
$$
\boxed{\begin{aligned}
(\pm1,0):&\quad f=\frac72\ln3,\\
(0,0):&\quad f=1+\frac72\ln3,\\
(0,\pm1/\sqrt2):&\quad f=\frac14+\frac72\ln\frac72.
\end{aligned}}
$$
The cases $x=0$ and $s=1$ exhaust the first <partial derivative> equation, so no other <stationary points> have been lost by division.
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