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? 

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.
