This question requires us to perform a technique called completing the square, which (in my opinion) is skirting the limits of what the GRE tests

That said here's the solution....

Take: $-3x^2 + 12x -2y^2 - 12y -39$

Rewrite as follows: $-3(x^2 + 4x) - 2(y^2 + 6y) -39$

Aside: At this point we need to recognize that $x^2 + 4x$ will become a perfect square if we add 4, because $x^2 + 4x + 4 = (x+2)(x+2) = (x+2)^2$
Similarly $y^2 + 6y + 9 = (y + 3)(y + 3) = (y + 3)^2$

So I'm going to add the following to the algebraic expression: $-3(x^2 + 4x + 4 - 4) - 2(y^2 + 6y + 9 - 9) -39$
All I've done here is add and subtract 4 from the first quantity, and add and subtract 9 from the second quantity. Since the net effect is simply adding 0 the both quantities, this is a legitimate step

Now remove the $-4$ and the $-9$ from both brackets (they're not needed anymore)
Keep in mind that removing $-4$ from the first set of brackets requires us to first multiply $-4$ by $-3$, removing $-9$ from the second set of brackets requires us to first multiply $-9$ by $-2$

When we do this we get: $-3(x^2 + 4x + 4) + 12 - 2(y^2 + 6y + 9) + 18 -39$

Simplify: $-3(x^2 + 4x + 4) - 2(y^2 + 6y + 9) - 9$

Factor: $-3(x+2)(x+2) - 2(y+3)(y+3) - 9$

Rewrite as follows: $-3(x+2)^2 - 2(y+3)^2 - 9$

At this point we need to recognize that $-3(x+2)^2$ will be less than or equal to 0 for all values of x
So, the greatest the value of $-3(x+2)^2$ is ZERO, and this occurs when $x = -2$

Likewise,$-2(y+3)^2$ will be less than or equal to 0 for all values of y
So, the greatest the value of $-2(y+3)^2$ is ZERO, and this occurs when $y = -3$

So, the greatest value of $-3(x+2)^2 - 2(y+3)^2 - 9$ occurs when $x = -2$ and $y = -3$
When we plug $x = -2$ and $y = -3$ into the equation, we get $-9$

So the greatest possible value is $-9$

Answer: B