If $\int f(x) d x=2\{f(x)\}^3+C$, then $f(x)$ is

$\sqrt{\frac{x}{3}}$

We have,

$\int f(x) d x=2\{f(x)\}^3+C$

Differentiating both sides w.r. to, $x$, we get

$f(x)=6\{f(x)\}^2 f'(x)$

$\Rightarrow 6 f(x) f'(x)=1$

$\Rightarrow 6 \int f(x) f'(x) d x=\int 1 d x$

$\Rightarrow 6 \int f(x) d(f(x))=\int 1 . d x$ 

$\Rightarrow 6 \times \frac{\{f(x)\}^2}{2}=x \Rightarrow\{f(x)\}^2=\frac{x}{3} \Rightarrow f(x)=\sqrt{\frac{x}{3}}$