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