Trigonometric substitution perhaps?

Indeed, it is possible to solve this problem by completing the square.

First, we express ${ x }^{ 2 }+4\sqrt { 3 } xy+{ 6y }^{ 2 }$ as ${ x }^{ 2 }+4\sqrt { 3 } xy+{ 6y }^{ 2 } +k(x^2-y^2)-2k$ .

This is equivalent to $(k+1)(x^2) + (6-k)(y^2) + 4\sqrt{3} xy -2k$ .

To complete the square, $\sqrt{k+1} \sqrt{6-k}=2\sqrt 3$ .

In this case $\sqrt{(k+1)(6-k)}=2\sqrt 3$ as it is impossible for both $k+1$ and $6-k$ to be negative.

Solving this gives $k=2$ or $k=3$ .

If $k=2$ , the expression would be $3x^{2}+4\sqrt{3}xy+4y^2-4$

$3x^{2}+4\sqrt{3}xy+4y^{2}-4=(\sqrt3x+2y)^2-4$

If $k=3$ , the expression would be $4x^{2}+4\sqrt{3}xy+3y^2-6$ .

$4x^{2}+4\sqrt{3}xy+3y^2-6=(2x+\sqrt3y)^2-6$ .

So the minimum value is supposed to be $-6$ , when $k=3$ , but we still need to check for equality case.

Equality holds when

$\begin{cases} 2x+\sqrt { 3 } y=0 \\ { x }^{ 2 }-{ y }^{ 2 }=2 \end{cases}$ .

$\Longrightarrow x=\frac { -\sqrt { 3 } y }{ 2 }$

$x^2=\frac{3y^2}{4}$

$x^2-y^2=\frac{-y^2}{4}=2$ , which has no real solutions for $y$ .

Hence, the minimum possible value of the expression CANNOT be $-6$ , as it is given in the question that the variables are real.

Now, the minimum value should be $-4$ , when $k=2$ , but we still need to check for equality case.

Equality holds when

$\begin{cases} \sqrt { 3 } x+2y=0 \\ { x }^{ 2 }-{ y }^{ 2 }=2 \end{cases}$ .

Solving this system of equations gives $\begin{cases} x=2\sqrt { 2 } ,\quad y=-6 \\ x=-2\sqrt { 2 } ,\quad y=6 \end{cases}$ .

So the minimum value of the expression is $-4$ , as it is given in the question that the variables are real.