Problem 5. Grouping the terms, we get: $\left(4 y^{2}+1\right)\left(x^{4}+2 x^{2}+2\right)=8|y|\left(x^{2}+1\right)$.

Obviously, there are no solutions when $y=0$. Dividing both sides by $2|y|\left(x^{2}+1\right)$:

$$
\left(2|y|+\frac{1}{2|y|}\right)\left(x^{2}+1+\frac{1}{x^{2}+1}\right)=4
$$

It is known ${ }^{1}$ (easy to show) that if $a>0$, then $a+\frac{1}{a} \geq 2$, and equality is achieved only if $a=1$. Therefore, the right side equals 4 only if $2|y|=1$ and $x^{2}+1=1$.
Answer: $\left(0, \frac{1}{2}\right),\left(0,-\frac{1}{2}\right)$.