If $f\left(x\right)=\int\limits _{0}^{x}f\left(t\right)dt$ then $f \equiv 0$?

I think there is a simpler method for that... As far as I know, there is only function satisfying $F(x)=f(x)$ (where F is an antiderivative of f). This function is $f(x)=ae^x$ by definition of the exponential function. Giving your situation, $F(x) = 0$ for $x=0$ but $\forall x\in\Bbb R, e^x > 0$ so $a=0$. Then the only possible function $f$ satisfying $F(x)=f(x)$ is $f(x)=0$.

EDIT : you can also think using the air representation of the integral. $\int\limits_a^b f(x)dx$ is, when $f(x)$ is continuous on $[a;b]$, the algebric measure of the air bounded by $x=a, x=b, y=0$ and $y=f(x)$. So when you write $\int\limits_0^x f(t)dt=0$, this means that the air is null for all $x \in\Bbb R$. Then the only function having a null air over the reals is $f(x)=0$.