Construct F(x) = f(x)g'(x)-f'(x)g(x)

Then F(x) is derivable at (a, b), and F(a) = F(b)=0.

f '(x)= f '(x)g '(x)+f(x)g ' '(x)-[f ' '(x)g(x)+f '(x)g '(x)]

= f(x)g''(x) - f''(x)g(x)

According to Rolle's theorem, n∈(a, b) exists.

So F'(n) =0.

That is, f(n)g''(n)-f''(n)g(n) =0.

That is, f (n)/g (n) = f "(n)/g" (n)

The original proposition was proved.

Note: The answer comes from netizen tian27546 Xixi.