3. Find all pairs of natural numbers $(x, y)$ such that $x y^{2}+1 \mid x^{2}+y^{3}$.

 untranslated text remains the same as requested.
3. Answer: they are the pairs $\left(t^{5}, t\right),\left(t, t^{3}\right),\left(t^{8}-t, t^{3}\right)$ for $t \in \mathbb{N}$ (and $t \geqslant 2$ in the last mentioned pair), as well as $(7,4)$ and (9,4).

Let's write $x^{2}+y^{3}=k x y^{2}+k$. For $x>k y^{2}$ we have $k-y^{3}=x\left(x-k y^{2}\right) \geqslant\left(k y^{2}+1\right)$. $1>k$, which is impossible. It follows that $x \leqslant k y^{2}$. For $x=k y^{2}$ we have (from the previous equation) $k=y^{3}$ and then $x=y^{5}$; the pair $\left(t^{5}, t\right)$ for $t \in \mathbb{N}$ is indeed a solution.

We further consider $xy^{2}$ :

Then $(k d)^{2}$ is a perfect square greater than $\left(k^{2} y^{2}-2 y\right)^{2}$ and of the same parity, so we have $\left(k^{2} y^{2}-2 y\right)^{2}+4\left(k^{3}-y^{2}\right)=(k d)^{2} \geqslant\left(k^{2} y^{2}-2 y+2\right)^{2}$, from which we get $k^{3} \geqslant\left(k^{2} y^{2}-2 y+1\right) \geqslant k^{2} y^{2}$. However, then we have $k \geqslant y^{2}$, which is impossible due to the earlier $k<y$.

- $k^{3}<y^{2}$ :

Then $(k d)^{2}$ is a perfect square less than $\left(k^{2} y^{2}-2 y\right)^{2}$ and of the same parity, so we have $\left(k^{2} y^{2}-2 y\right)^{2}+4\left(k^{3}-y^{2}\right)=(k d)^{2} \leqslant\left(k^{2} y^{2}-2 y-2\right)^{2}$, which is equivalent to $k^{3}+\left(k^{2}-1\right) y^{2} \leqslant 2 y+1$. This is possible only for $k=1$. Now we have $d^{2}=y^{4}-4 y^{3}+4=\left(y^{2}-2 y-2\right)^{2}-8 y$, so from $d^{2} \leqslant\left(y^{2}-2 y-4\right)^{2}$ we get $y^{2}-4 y-3 \leqslant 0$. We obtain $y \leqslant 4$ (and $y \geqslant 2$, due to $1=k^{3}<y^{2}$). For $y=2$ we have $d^{2}=16-32+4=-28$, which is impossible. For $y=3$ we have $d^{2}=$ $81-108+4=-23$, which is impossible. For $y=4$ we have $d^{2}=256-256+4=4$, so we find the solutions $x=\frac{16 \pm \sqrt{4}}{2}$, i.e., $x=7$ or $x=9$.