1. We are given $3\int_1^x f(t)\,dt = x f(x) - \frac{x^3}{3}, \qquad x\in[1,\infty),$ with $f(1)=\frac13.$ We need to find $f(e)$.

2. Differentiate both sides with respect to $x$.

Using the Fundamental Theorem of Calculus, $\frac{d}{dx}\left(3\int_1^x f(t)\,dt\right)=3f(x).$

For the right-hand side, $\frac{d}{dx}\left(xf(x)-\frac{x^3}{3}\right)=f(x)+x f'(x)-x^2.$

Hence, $3f(x)=f(x)+x f'(x)-x^2.$ So, $2f(x)=x f'(x)-x^2,$ or $x f'(x)-2f(x)=x^2.$

3. Rewrite this as a linear differential equation: $f'(x)-\frac{2}{x}f(x)=x.$

4. Solve using integrating factor.

The integrating factor is $\mathrm{IF}=e^{\int -2/x\,dx}=e^{-2\ln x}=x^{-2}$ (since $x\ge 1$, so $x>0$).

Multiplying the equation by $x^{-2}$, $x^{-2}f'(x)-2x^{-3}f(x)=x^{-1}.$

The left side is $\frac{d}{dx}\left(f(x)x^{-2}\right)=\frac{1}{x}.$

Integrating, $f(x)x^{-2}=\ln x + C.$ Therefore, $f(x)=x^2(\ln x + C).$

5. Use the initial condition $f(1)=\frac13$.

Since $\ln 1=0$, $f(1)=1^2(0+C)=C=\frac13.$ So, $f(x)=x^2\left(\ln x+\frac13\right).$

6. Now compute $f(e)$.

Since $\ln e=1$, $f(e)=e^2\left(1+\frac13\right)=e^2\cdot \frac43=\frac{4e^2}{3}.$

7. Compare with options:

• A: $\frac{e^2+4}{3}$
• B: $\frac{\log_e 4 + e}{3}$
• C: $\frac{4e^2}{3}$
• D: $\frac{e^2-4}{3}$

Hence the correct option is $\boxed{\text{C}}.$