It's not 1 and 1

We have the following system of equations

$\begin{cases} x^2+5xy+y^2=7 \qquad\qquad (1)\\ x^2y+xy^2=2 \qquad\qquad \phantom{ccci} (2) \end{cases}$

From $(2)$ we get

$xy(x+y)=2 \iff x+y=\dfrac{2}{xy}$

Manipulating $(1)$ :

$\begin{aligned} x^2+5xy+y^2=7 &\iff x^2+2xy+y^2=7-3xy\\ &\iff (x+y)^2=7-3xy \end{aligned}$

Substituting what we get from $(2)$

$\begin{aligned} \left(\dfrac{2}{xy}\right)^2=7-3xy &\iff \dfrac{4}{x^2y^2}=7-3xy\\ &\iff 3x^3y^3-7x^2y^2+4=0 \end{aligned}$

A cubic in $xy$ which can be factored as

$(xy-1)(xy-2)\left(xy+\dfrac{2}{3}\right)=0$

Which implies $xy\in\left\{1,2,-\dfrac{2}{3}\right\}$

$xy=1$ can be immediately discarded, since

$x+y=\dfrac{2}{xy}=2$

But $x+y\neq 2$ from the problem hypotesis

$xy=\mathbin{\color{#D61F06}2}$ implies $x+y=\mathbin{\color{#3D99F6}1}$ , so $x$ and $y$ are the solutions of the equation

$t^2-\mathbin{\color{#3D99F6}1}t+\mathbin{\color{#D61F06}2}=0$

Which has no real solutions. $xy=\mathbin{\color{#D61F06}-\dfrac{2}{3}}$ implies $x+y=\mathbin{\color{#3D99F6}-3}$ , so $x$ and $y$ are the solutions of

$t^2-\mathbin{\color{#3D99F6}(-3)}t+\mathbin{\color{#D61F06}-\dfrac{2}{3}}=0\iff 3t^2+9t-2=0$

Which do have real solutions, thus $xy=-\dfrac{2}{3}$ . To conclude

$(6xy)^2=\left[6\left(-\dfrac{2}{3}\right)\right]^2=(-4)^2=\boxed{16}$