Let f(x)=g(x)h(x), g(10)=-4, h(10)=560, g'(10)=0, and h'(10)=35. find f'(10)

This question has extraneous info to trick you.
f(x) = g(x)h(x) ⇒ f'(x) = g'(x)h'(x) . Letting x = 10, we get f'(10) = g'(10)h'(10). then just plug in the values provided. g(10) and h(10) are there to throw you off, just use g'(10) and h'(10).
So f'(10), pronounced "eff prime of ten", = 0 * 35 = 0.

If the question were asking for f(10) instead of f'(10) then you would use g(10) and h(10), ⇒-4*560=90.