(fog)'(x) = f'(g(x))*g'(x))
If you can adidquately define (fog)'(3), I can answer the question. I'm not sure what to do with the prime. Does that mean 'take the derivative of (fog)' or does that mean 'the derivative of f OF the derivative of g'? Or does it have nothing to do with derivatives at all? Until I know which is the case, there's not much I can do to help.
(fog)'(3) = f'(g(3))g'(3)
g(3) = 4 ... just go to x = 3, then move up until you hit the graph of g(x), that y value is the value of g(3), which is 4.
f'(g(3)) = f'(4) = 2 ... go to x = 4, then up to the graph of f(x), estimate the slope of f(x) at that point and you will have the value of f'(4), which is 2.
g'(3) = -1 ... explanation: same as above
(fog)'(3) = (2)(-1) = -2
(gof)'(3) = g'(f(3))f'(3)
f(3) = 0
g'(f(3)) = g'(0) = 0
f'(3) = 2
(gof)'(3) = (0)(2) = 0