\(P=x^2y^2+x^2-2xy+6x+2013\)
\(P=\left(xy-1\right)^2+\left(x^2+6x+9\right)+2003=\left(xy-1\right)^2+\left(x+3\right)^2+2003\ge2003\)
\(\Rightarrow Min_P=2003\Leftrightarrow\hept{\begin{cases}xy=1\\x+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-\frac{1}{3}\\x=-3\end{cases}}\)