Can we minimize it?

Let $x=r\cos t,y=r\sin t$

Then $r^4\cos t\sin t (\cos^2t-\sin^2t)=r^2,r^2\times\dfrac{\sin 2t}2\times\cos 2t=1,r^2\times\dfrac{\sin 4t}4=1$

We are looking for the minimum of $r^2=\dfrac4{\sin 4t}$

Of course,the minimum happens when $\sin 4t=1,r^2=4$

Let $x^2=p$ , $y^2={q}$

So we have $\large xy(p-q)=p+q$

Squaring both sides,

$\large pq(p-q)^2=(p+q)^2$

$\large\Rightarrow pq(p+q)^2-4(pq)^2=(p+q)^2$

$\large\Rightarrow (pq-1)(p+q)^2=4(pq)^2$

$\large\Rightarrow (p+q)^2=\frac{4(pq)^2}{pq-1}$

Therefore $\large\min{[(p+q)^2]}=\min{[\frac{4(pq)^2}{pq-1}}]$

Now we know, $\large(pq-2)^2\geq0$

$\large\Rightarrow (pq)^2-4pq+4\geq0$

$\large\Rightarrow (pq)^2\geq4(pq-1)$

$\large\Rightarrow \frac{(pq)^2}{pq-1}\geq4$

$\large\Rightarrow \frac{4(pq)^2}{pq-1}\geq16$

$\large\Rightarrow (p+q)^2\geq16$

$\large\Rightarrow (p+q)\geq4$

$\large\Rightarrow (x^2+y^2)\geq\boxed{4}$