Answer
False
Work Step by Step
Counter-example:
$f(x)=(x-1)(x+1)\rightarrow g(x)=(x-1)$; $h(x)=(x+1)$
$g'(x)=1$; $h'(x)=1\rightarrow$ If it was true $f'(x)=1$
But by expanding we see that $f(x)=x^2-1\rightarrow f'(x)=2x.$
The rule is therefore false. Also, it contradicts the product rule.
