I wish to determine all functions $f$ of bounded variation on $[0, 1]$ such that $$ f(x) +(TV(f_{[0,x]}))^{1/2} = 1 \quad \text{for all } x \in [0,1] \quad \text{and} \quad \int_0^1 f(x) \: dx = 1/3.$$
Here $TV(f_{[a,b]})$ denotes the total variation of $f$ on $[a,b]$.
So looking at what needs to be done, it is clear to me that the function f(x) = 1/3 satisfies the integral equation however, I have no intuition on how to proceed with the equation that involves the total variation or how to bring it all together.
In these cases it is a good idea to start plugging particular values and see what happens: here $x=0$ tells us that $f(0) = 1$, because $f_{[0,0]} $ has $0$ variation.
Another useful fact is that $TV(f_{[0,a]}) \le TV(f_{[0,b]})$ for $a \le b$. Then, from your first equation you have $f(a)= 1 - TV(f_{[0,a]}) \ge 1 - TV(f_{[0,b]}) \ge f(b)$, so $f$ is non-increasing.
For non-increasing functions, the total variation simplifies to $TV(f_{[0,x]}) = f(0) - f(x) = 1 - f(x) $ and your first equation becomes $$ (1 - f(x))^{1/2} = 1 - f(x), $$ which tells us that $f(x) \in \{0, 1\}$ for all $x \in [0,1]$.
EDIT: Since $f$ is non-increasing and is either $0$ or $1$, the only way to satisfy the integral equation is having $f(x) = 1, \forall x<1/3$ and $f(x)=0, \forall x>1/3 $. The value at $x = 1/3$ does not change the integral, so both $f(1/3)=0$ and $f(1/3)=1$ are possible. So your problem has $2$ solutions.