As is always true, with "nice" functions, the two mixed derivatives are equal. You could find [itex]\partial^2 f/\partial x\partial y[/itex] by differentiating first with respect to x, then with respect to y: first getting -6x and then, since it does not depend on y, 0. Or you could differentiate first with respect to y, then with respect to x: getting 0 immediately and then, of course, the derivative of "0" with respect o x is 0.
I, and I suspect many who read your post, was momentarily taken aback since I thought you were "holding x constant" through both derivatives. But you are correct: since this function does not depend on y, any derivative of it with respect to y, is 0.