A clever manipulation

$\begin{aligned} 2x^2 + y^2 - 2xy + 12y + 72 & \le 0 \\ 2x^2 - 2xy + (y+6)^2 + 36 & \le 0 \\ \blue{x^2} + \red{x^2 - 2x(y+6)} \blue{+ 12x} \red{+ (y+6)^2} \blue{+ 36} & \le 0 \\ \blue{(x+6)^2} + \red{(x-(y+6))^2} & \le 0 \end{aligned}$

We note that the $\text{LHS} \ge 0$ . Hence there is only one solution that is the $\text{LHS} = (x+6)^2 + (x-(y+6))=0$ , which is only true when $x + 6 = 0$ and $x-(y+6)=0$ . Therefore, $x=-6$ , $y=-12$ , and $x^2y = \boxed{-432}$ .

Multiply the equation by 2:

$\implies 4x^2+2y^2-4xy+24y+144\leq 0$

$\implies (4x^2+y^2-4xy)+(y^2+24y+144)\leq 0$

$\implies (2x-y)^2+(y+12)^2\leq 0$

Since the sum of squares are always non-negative, we must have $2x-y=0$ and $y-12=0$ . Hence $x=-6$ and $y=-12$ .

This gives us $x^2y=(-6)^2\cdot(-12)=\boxed{-432}$ .

say $x=p$ and $y = p + q$ . We have:

$2p^2 + (p+q)^2-2p(p+q)+12(p+q)+72\leq 0$

$p^2+q^2+12p+12q+72\leq 0$

$p^2+12p+36+q^2+12q+36\leq 0$

$(p+6)^2+(q+6)^2\leq 0$

Now, we can see $p=-6$ and $q=-6$ . Which gives $x=-6$ and $y=-12$

Finally, $x^2y=-12 \cdot 36 = -432$