The integral equation is: $$\int_0^x f(t)\,dt = 3f(x)+k,\quad \quad(\star)$$ where $k$ is a constant. Differentiating both sides of $(\star)$ gives $$f(x) = 3f'(x)$$ If there is a solution of $(\star)$ it must be of the form $$f(x) = Ae^{x/3},$$ for some constant $A$. We check to see whether or not this is a solution.
For $f(x)=Ae^{x/3}$ we have $$\int_0^x Ae^{t/3}\,dt = \Big[3Ae^{t/3}\Big]_0^x = 3Ae^{x/3}-3A.$$ Thus $f(x)=Ae^{x/3}$ is a solution if and only if $A=-k/3$. The unique solution is $$f(x) = {-k\over 3} e^{x/3}.$$