Prove/disprove: $f+g$ and $f$ are differentiable at $x_0$ $\implies$ $g$ is differentiable at $x_0$ 
Prove/disprove: $f+g$ and $f$ are differentiable at $x_0$ $\implies$ $g$ is differentiable at $x_0$  

attempt 
Suppose $g$ is not differentiable at $x_0$. There are three cases:
Case I: The two following one-sided limits exist, but not equal.$$\lim_{x\to {x_0}^+}\frac{g(x)-g(x_0)}{x-x_0}\neq \lim_{x\to {x_0}^-}\frac{g(x)-g(x_0)}{x-x_0}$$Case II: $g$ is not continuous at $x_0$
Case III: $g$ is undefined at $x_0$  
Case I: $$\lim_{x\to {x_0}^-}\frac{(f+g)(x)-(f+g)(x_0)}{x-x_0}=\lim_{x\to {x_0}^-}\Big(\frac{f(x)-f(x_0)}{x-x_0}+\frac{g(x)-g(x_0)}{x-x_0}\Big)=\lim_{x\to {x_0}^-}\frac{f(x)-f(x_0)}{x-x_0}+\lim_{x\to {x_0}^-}\frac{g(x)-g(x_0)}{x-x_0}=L_f+L_{g^-}$$Similarly, with the other one-sided limit:$$\lim_{x\to {x_0}^+}\Big(...\Big)=...=L_f+L_{g^+}$$  
Case II: For every type of discontinuity of $g$ we use one-sided limit rules and continuity of $f$ to show that $f+g$ is not continuous and therefore not differentiable.   
Case III: That leads to $f+g$ being undefined at $x_0$.  
comment
I haven't yet found a counterexample. If this idea of a proof works, is there a shorter one?

you can  prove that if $ h_1$ and $h_2$ are differntiable than $h_1+h_2$ is differentiable and that if $h$ is differentiable than $c\cdot h$ is differentiable 
(So the set of differentiable function in one point $x_0$ is a subspace of the Vectorial space of the continuos function in $x_0$)
In this case $g=(f+g)-f$ and so you have finished
