1. We are given a relationship between the integral of the function $f(t)$ and the function $f(x)$ :

$$3 \int\limits_1^x f(t) d t=x f(x)-\frac{x^3}{3}, x \in[1, \infty)$$

2. Differentiate both sides of the above equation with respect to $x$ :

$$ 3(f(x) \cdot 1-0)=1 . f(x)+x f^{\prime}(x)-\frac{3 x^2}{3} $$

$$ \Rightarrow $$ $$3f(x) = f(x) + xf'(x) - x^2$$

$$ \Rightarrow $$ $$2f(x) = xf'(x) - x^2$$

3. We can rewrite this as a first order linear differential equation:

$$xf'(x) - 2f(x) = x^2$$

4. The general form of a first order linear differential equation is:

$$y'(x) + p(x)y = q(x)$$

Here $y = f(x)$, $p(x) = -\frac{2}{x}$, and $q(x) = x$.

5. The integrating factor is obtained by exponentiating the integral of $p(x)$ with respect to $x$:

$$e^{\int p(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2\ln|x|} = \frac{1}{x^2}$$

6. $$ \therefore $$ Solution of D.E :

$$ \begin{gathered} y \cdot\left(\frac{1}{x^2}\right)=\int(x) \frac{1}{x^2} d x \\\\ \frac{y}{x^2}=\ln x+c \\\\ y=x^2 ln x+c x^2 \end{gathered} $$

$$ \therefore $$ The general solution is :

$$y(x) = x^2(\ln x + c)$$

Here, $y(x)$ is the same as $f(x)$, i.e. the function we're trying to find.

7. Given that $f(1) = \frac{1}{3}$, let's use this condition to find the constant $c$ :

$$\frac{1}{3} = 1^2(\ln 1 + c) \Rightarrow c = \frac{1}{3}$$

Therefore, the function $f(x)$ we are looking for is:

$$f(x) = x^2(\ln x + \frac{1}{3})$$

We want to find $f(e)$, so substituting $x=e$ into the function, we get :

$$f(e) = e^2(\ln e + \frac{1}{3}) = e^2(1 + \frac{1}{3}) = \frac{4e^2}{3}$$

So, the correct answer is

Option C : $\frac{4e^2}{3}$