Find if the function $\frac{(1-2xy)}{(x^2 +y^2)}$ has a max or min value for $(x,y)=/=(0,0)$

Does the function $$\frac{1-2xy}{x^2 +y^2}$$ have a max or min value for $$(x,y)=/=0$$?

What I've tried so far is to take the the partial derivatives:

$$\frac{\partial f}{\partial x} = \frac{2(-x+x^2*y - y^3)}{(x^2 + y^2)^2}$$ $$\frac{\partial f}{\partial y} = \frac{2(x^3 -x*y^2 +y)}{(x^2 + y^2)^2}$$

However I can't see what satisfy will $$\nabla f(a,b)=0$$

It looks like the function has a singular point in $$(0,0)$$ since it doesn't exist there, but I am told to ignore that point.

And seeing there is no boundary to f, the max/min can't be there either.

So, how can I then show that this function has a max/min other than in $$(0,0)$$?

In polar coordinate you get $$f(x,y)=\dfrac{1-2xy}{x^2+y^2}=\dfrac {1-2r^2\sin(\theta)\cos(\theta)}{r^2}=\dfrac 1{r^2}-\sin(2\theta)$$
So since $$\dfrac 1{r^2}>0$$ and $$\sin(2\theta)\in[-1,1]$$ we have a lower bound $$-1$$.
The minimum cannot be reached though because even if $$-\sin(2\theta)$$ has a minimum $$-1$$ along the line $$(2\theta)=\frac{\pi}2\iff y=x$$ the part in $$\dfrac 1{r^2}$$ has no mininum, it decreases to zero at infinity.
Indeed $$f(x,x)=\dfrac 1{2x^2}-1$$ doesn't reach its lower bound.
As for a maximum, the upper bound is $$+\infty$$ because it is dominated by $$\dfrac 1{r^2}$$ when $$r\to 0$$, so there is no maximum either.
Hint: we get $$\frac{1-2xy}{x^2+y^2}>-1$$ since we have $$(x-y)^2+1>0$$