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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 $(0,0)$?

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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.

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Hint: we get $$\frac{1-2xy}{x^2+y^2}>-1$$ since we have $$(x-y)^2+1>0$$

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