6.076. $\left\{\begin{array}{l}x^{2} y^{3}+x^{3} y^{2}=12 \\ x^{2} y^{3}-x^{3} y^{2}=4\end{array}\right.$

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6.076. $\left\{\begin{array}{l}x^{2} y^{3}+x^{3} y^{2}=12 \\ x^{2} y^{3}-x^{3} y^{2}=4\end{array}\right.$
Solution.

From the condition we have $\left\{\begin{array}{l}x^{2} y^{2}(y+x)=12, \\ x^{2} y^{2}(y-x)=4 .\end{array}\right.$

Dividing the first equation by the second, we get

$\frac{x^{2} y^{2}(y+x)}{x^{2} y^{2}(y-x)}=\frac{12}{4} \Leftrightarrow \frac{y+x}{y-x}=3 \Leftrightarrow y+x=3 y-3 x \Leftrightarrow y=2 x$.

From the first equation of the system, we find

$x^{2}(2 x)^{3}+x^{3}(2 x)^{2}=12 \Leftrightarrow 8 x^{5}+4 x^{5}=12 \Leftrightarrow x^{5}=1, x=1$.

Then $y=2$.

Answer: $(1 ; 2)$