Intermediate value theorem: showing there exists $c \in [0,2/3]$ such that $f(x+1/3) = f(x)$ Intermediate value theorem: show there exists $c \in [0,2/3]$ such that $f(c+\frac{1}{3}) = f(c)$
Let $f: [0,1] \to [0,1]$ be continuous and $f(0) = f(1)$
This is pretty straight forward and the other ones ive done it was easier to find why IVT applies, namely find why it changes signs.
Letting $g(x) = f(x + \frac{1}{3}) - f(x)$ then this is only defined for the interval $[0,\frac{2}{3}]$ b/c not sure what $g(1) = f(\frac{4}{3}
) - f(1)$ may be.
Now, $g(0) = f(\frac{1}{3}) - f(0)$
and $g(\frac{2}{3}) = f(1) - f(\frac{2}{3}) = f(0) - f(\frac{2}{3})$. 
I can't (at least I don't see why I can) conclude this is equal to $-g(0)$ which has been the case in other exercises, which is what I used to apply the IVT and get the wanted result. 

HINT:
You were on the right track by defining $g(x)\equiv f(x+1/3)-f(x)$ for $x\in [0,2/3]$.  
Now, note that $$g(0)+g(1/3)+g(2/3)=f(1)-f(0)=0$$Then either $g(0)=g(1/3)=g(2/3)=0$ or at least one of the three is positive and one of the three is negative.  Now, use the IVT.
