System of differential equations, compute value 
Suppose $f(x)$ and $g(x)$ are nonconstant smooth functions satisfying the equations $$\frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)} = 1$$ $$ \frac{f''(x)}{f(x)} + \frac{g''(x)}{g(x)} = \frac{f'''(x)}{f(x)} + \frac{g'''(x)}{g(x)}$$ (assume $f$ and $g$ are always nonzero). Compute $$ \frac{f''(2015)}{f'(2015)} + \frac{g''(2015)}{g'(2015)}.$$

Evidently, rearranging the first condition and using product rule tells us $fg = Ce^x$ for some constant $C$. But I don't know how to proceed from here.
 A: You have
$$ \begin{align}
\frac{f'}{f} + \frac{g'}{g} &= 1 \\\\
\implies (fg)' &= fg \\
\end{align} $$
Solving, you found 
$$ fg = Ce^{x} $$
Now, taking $\frac{d}{dx}$ of both sides of your first equation gives
$$ \begin{align}
\frac{d}{dx} \bigg ( \frac{f'}{f} + \frac{g'}{g} \bigg) &= \frac{d}{dx} ( 1 ) \\\\
\implies \frac{f''}{f} - \frac{f'^{2}}{f^{2}} + \frac{g''}{g} - \frac{g'^{2}}{g^{2}} &= 0 \\
\implies \frac{f''}{f} + \frac{g''}{g} &= \frac{f'^{2}}{f^{2}} + \frac{g'^{2}}{g^{2}} \\
\implies \frac{f''}{f} + \frac{g''}{g} &= \bigg (\frac{f'}{f} + \frac{g'}{g} \bigg )^{2} - 2 \frac{(fg)'}{fg} \\
&= -1
\end{align} $$
where we solved the RHS using the earlier results. Does this help?
A: i believe the answer is  ${2 \over 3}$ 
we are given $fg^\prime + fg^\prime = fg, \ 
f^{\prime \prime}g+fg^{\prime \prime} = f^{\prime \prime \prime}g+fg^{\prime \prime \prime}$
differentiating twice $(fg)^\prime = fg^\prime + f^\prime g = fg  $ 
$$fg^{\prime \prime} + f^{\prime \prime}g + 2f^\prime g^\prime = fg$$
$$fg^{\prime \prime\prime} + f^{\prime \prime \prime}g + 3f^{\prime \prime} g^\prime   +3f^{\prime} g^{\prime \prime}= fg$$
we get $${ 3(f^{\prime \prime} g^\prime   +3f^{\prime} g^{\prime \prime})  
\over 2f^\prime g^\prime  }  = { fg - (fg^{\prime \prime\prime} + f^{\prime \prime \prime}g) \over fg - (fg^{\prime \prime} + f^{\prime \prime}g ) } = 1$$
so that $$ { f^{\prime \prime} g^\prime   +f^{\prime} g^{\prime \prime}  
\over f^\prime g^\prime  } = {f^{\prime \prime} \over f^\prime} + {g^{\prime \prime} \over g^\prime} = {2 \over 3}  $$ 
