3.6. $\left\{\begin{array}{l}x^{3}-y^{3}=26, \\ x^{2} y-x y^{2}=6 .\end{array}\right.$
3.6. Let $y=k x$. Immediately note that $k \neq 1$. From the equations

$$
\begin{aligned}
& x^{3}-k^{3} x^{3}=26 \\
& k x^{3} y-k^{2} x^{3}=6
\end{aligned}
$$

we get $x^{3}=\frac{26}{1-k^{3}}$ and $x^{3}=\frac{6}{k-k^{2}}$. Therefore,

$$
\frac{26}{1-k^{3}}=\frac{6}{k-k^{2}}
$$

This equation can be multiplied by $1-k$. As a result, we get

$$
\frac{26}{1+k+k^{2}}=\frac{6}{k}
$$

from which $k=3$ or $1 / 3$. Therefore, $x^{3}=-1$ or $x^{3}=27$. In the end, we get the following solutions: $(-1,-3),\left(\frac{1 \pm i \sqrt{3}}{2}, \frac{3}{2}(1 \pm i \sqrt{3})\right),(3,1)$, $\left(\frac{-3 \pm 3 i \sqrt{3}}{2}, \frac{1}{2}(-1 \pm i \sqrt{3})\right)$.