Showing that $f(x,y)=(\lvert x\rvert -\lvert y\rvert)\log\left(2x^2+\lvert y\rvert\right)$ has no local extrema I have an exercise asking to find the local extrema of $$ f(x,y)=(\lvert x\rvert -\lvert y\rvert)\log\left(2x^2+\lvert y\rvert\right)$$if they exist... Wolfram Alpha tells me they don't, but I think I've spent too long trying to show it, so I'm wondering what I missed, resulting in a more time-consuming task. Below is my work:
$f$ is continuous on the whole plane except the origin. 
For $x,y\ne0$$$f(x,y)=\begin{cases}(x-y)\log\left(2x^2+y\right) &x,y>0 \\ (x+y)\log\left(2x^2-y\right) &x>0>y\\-(x+y)\log\left(2x^2+y\right) &x<0<y \\ -(x-y)\log\left(2x^2-y\right) &x,y<0\end{cases}$$hence $$f_x(x,y)=\begin{cases}\log\left(2x^2+y\right)+\dfrac{4x(x-y)}{2x^2+y} &x,y>0 \\ \log\left(2x^2-y\right)+\dfrac{4x(x+y)}{2x^2-y} &x>0>y \\ -\log\left(2x^2+y\right)-\dfrac{4x(x+y)}{2x^2+y} &x<0<y \\ -\log\left(2x^2-y\right)-\dfrac{4x(x-y)}{2x^2-y} &x,y<0 \end{cases} $$ and
$$f_y(x,y)=\begin{cases}-\log\left(2x^2+y\right)+\dfrac{x-y}{2x^2+y} &x,y>0 \\ \log\left(2x^2-y\right)-\dfrac{x+y}{2x^2-y} &x>0>y \\ -\log\left(2x^2+y\right)-\dfrac{x+y}{2x^2+y} &x<0<y \\ \log\left(2x^2-y\right)+\dfrac{x-y}{2x^2-y} &x,y<0 \end{cases}.$$ Setting both partial derivatives equal to $0$ I got $$\begin{cases}(x-y)(4x+1) &x,y>0 \\ (x-y)(4x+1) &x>0>y \\ (x+y)(4x-1) &x<0<y \\ (x-y)(4x-1) &x,y<0 \end{cases} \iff y=\pm x\ne0.$$ So for $x_0\ne0$ and small $h_1,h_2$ I examined \begin{align}\operatorname{sgn}\left\{f(x_o+h_1,\pm x_0+h_2)-f(x_0,\pm x_o)\right\}&:=\operatorname{sgn}(h_3\log(2x_0^2+|x_0|+h_4)) \\&= \operatorname{sgn}(h_3\log(2x_0^2+|x_0|)) \end{align}which obviously depends on how $h_3$ approaches $0$, so the $(x_0,\pm x_0)$ are saddle points.
EDIT: ok on second thought I forgot to show that $h_3$ doesn't depend on $h_1, h_2$ in a way that would make it approach $0$ only from the left/right, but let's pretend I did show it
Now I should study the behaviour around points $(x_0,0)$ and $(0,y_0)$ , and this is where I've been stuck. After a while I tried looking at things like $f(x_o,h)-f(x_0,0)$ but it didn't help... what am I missing? 
 A: In the following I am assuming that $(x,y) \neq (0,0)$.
Since $f(x,y) = f(|x|,|y|)$, we need only examine $x\ge 0, y \ge 0$.
For $x>0,y>0$ we have
$f_x(x,y) = {1 \over 2x^2+y} (4x (x-y) + (2x^2+y) \log(2x^2+y))$,
$f_y(x,y) = {1 \over 2x^2+y} ((x-y) - (2x^2+y) \log(2x^2+y))$.
Suppose $f_x(x,y) = f_y(x,y) = 0$, then the above gives
$x = - { 1 \over 4}$, hence $f$ has no stationary points in $x>0, y>0$.
The axes need more care, as $f$ is not differentiable there.
Now consider the $x$ axis, we have $f_x(x,0) = \log(2 x^2)+2$, which has
a
zero at $x^*={1 \over \sqrt{2}e }$. A little work shows that $x^*$ is a strict local $\min$ (of $x \mapsto f(x,0)$).
If we consider the $y \ge 0$ side of the graph, we have  $f_y(x^*,0) > 0$ (the one sided derivative) hence, with a little finesse, we can conclude that $f$ has a local $\min$ at $(x^*,0)$.
Now consider the $y$ axis, we have $f(0,y) = -y \log(y)$, which has a strict $\max$ at $y^* = {1 \over e}$. As above, by restricting to $x \ge 0$ and considering the one sided derivative, we see that
$f_x(0,y^*) = -1 <0$ from which we can conclude that $(0,y^*)$ is a strict
local $\max$.
To summarise, $f$ has strict local $\min$ at $(\pm {1 \over \sqrt{2} e}, 0)$ and
strict local $\max$ at $(0, \pm {1 \over e})$. There are no other
extrema.
A: The function 
$$
f(x,y) = \left(|x|-|y|\right)\log(2x^2-|y|)
$$
has symmetries along the $x$ and $y$ axis because
$$
f(x,y) = f(\pm x, \pm y)
$$
Now considering the first quadrant or
$$
f(x,y) =(x-y)\log(2x^2-y)
$$
making the change of variable $u = 2x^2+y$ we have
$$
g(x,u) = (x+2x^2-u)\log u
$$
in this coordinate system the stationary points are given as the solutions for
$$
\nabla g(x,u) = \left\{
\begin{array}{rcl}
(1+4x)\log u & = & 0\\
x+2x^2-u - u\log u & = & 0
\end{array}\right.
$$
and solving we have
$$
\left(
\begin{array}{ccc}
x & u & y \\
 -1 & 1 & -1 \\
 -\frac{1}{4} & -\frac{1}{8 W\left(-\frac{e}{8}\right)} & -\frac{1}{8}-\frac{1}{8 W\left(-\frac{e}{8}\right)} \\
 -\frac{1}{4} & -\frac{1}{8 W_{-1}\left(-\frac{e}{8}\right)} & -\frac{1}{8}-\frac{1}{8 W_{-1}\left(-\frac{e}{8}\right)} \\
 \frac{1}{2} & 1 & \frac{1}{2} \\
\end{array}
\right)
$$
Here $W(\cdot)$ is the so called Lambert function. In numbers
$$
\left(
\begin{array}{ccc}
x & u & y \\
 -1. & 1. & -1. \\
 -0.25 & 0.191569 & 0.0665694 \\
 -0.25 & 0.0860153 & -0.0389847 \\
 0.5 & 1. & 0.5 \\
\end{array}
\right)
$$
The feasible first quadrant solution gives
$$
x = \frac{1}{2}, u = 1, y = \frac{1}{2}
$$
The qualification for this stationary point is made with the contribution of the $g(x,u)$ hessian
$$
H_g = \left(
\begin{array}{cc}
 4 \log (u) & \frac{4 x+1}{u} \\
 \frac{4 x+1}{u} & -\frac{2 x^2+x-u}{z^2}-\frac{2}{u} \\
\end{array}
\right)
$$ 
At $(x,y) = (1/2,1/2)$ the hessian gives 
$$
H = \left(
\begin{array}{cc}
 0 & 3 \\
 3 & -2 \\
\end{array}
\right)
$$
characterizing this point as a saddle point. Considering the four quadrants the surface takes the form shown below.

and we can observe the existence of local minima undetected by the traditional method.

The minima are located at 
$$
\{x,y\} = \{\pm\frac{1}{e\sqrt2},0\}
$$
