Answer
-4
Work Step by Step
Recall that a parabola with vertex $(h, k)$ has a graph of the form $(y-k) = A(x-h)^2$ for some arbitrary constant A. Comparing with $f(x) = 2(x+4)^2 - 6$ (and with a little rearrangement), we can easily verify that our parabola's vertex is at $(-4, -6)$.
Any line of symmetry of the parabola must, obviously, pass through its vertex. Since this is an upward-opening parabola (the leading coefficient of x > 0), and it is symmetric about the y-axis, the line of symmetry must be a line parallel to the y-axis that passes through $(-4, -6)$. This can only be $x = -4$!