G'day mate,
"y" in this case is "a", I accidentally used "a" throughout, sorry!!!!!
\(\begin{bmatrix}\frac{1}{x}+\frac{1}{a}=\frac{1}{3}\\ xa+x+a=4\end{bmatrix}\)
\(\mathrm{Plug\:the\:solutions\:}x=-\frac{3a}{3-a}\mathrm{\:into\:}xa+x+a=4\)
We get a = \(a=\frac{1}{2}+i\frac{\sqrt{11}}{2} .and. a=\frac{1}{2}-i\frac{\sqrt{11}}{2}\)
\(\mathrm{Plug\:the\:solutions\:}a=\frac{1}{2}-i\frac{\sqrt{11}}{2},\:a=\frac{1}{2}+i\frac{\sqrt{11}}{2}\mathrm{\:into\:}\frac{1}{x}+\frac{1}{a}=\frac{1}{3}\)
That gives \(x=\frac{1}{2}+\frac{\sqrt{11}}{2}i and x=\frac{1}{2}-\frac{\sqrt{11}}{2}i\)
So now we have our values for (x, y): \(\begin{pmatrix}x=\frac{1}{2}+\frac{\sqrt{11}}{2}i,\:&y=\frac{1}{2}-i\frac{\sqrt{11}}{2}\\ x=\frac{1}{2}-\frac{\sqrt{11}}{2}i,\:&y=\frac{1}{2}+i\frac{\sqrt{11}}{2}\end{pmatrix}\)
Plugging that in: \(x^2\cdot y + x\cdot y^2\):
\(\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)^2\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)+\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)^2=3\)
\(\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)^2\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)+\left(\frac{1}{2}-\frac{\sqrt{11}}{2}i\right)\left(\frac{1}{2}+\frac{\sqrt{11}}{2}i\right)^2=3\)
So the answer is \(3.\)
-Vinculum