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.