I am given
$$ X(t) = (1+W(t)/3)^3 $$
and am asked to show that it solves
$$ dX = \frac{1}{3}X^{1/3}dt + X^{2/3}dW\\ X(0)=1 $$
By using Ito's rule on $X$, it is quite easy to arrive to $dX = \frac{1}{3}X^{1/3}dt + X^{2/3}dW$. So here I go from $X$ to $dX$.
However, this does not make use of the initial condition $X(0)=1$ in any way. In order to use the initial condition, I guess one has to go from $dX$ to $X$. How does one go about that?
In general, the differential expression of SDE's are shorthand for an integral equation. So you actually do have the expression for $X$ already. $$ \text{d}X_{t} = \frac{1}{3}X_{t}^{\frac{1}{3}}\text{d}t+X^{\frac{2}{3}}\text{d}W_{t} \iff X_{t} - X_{0} = \frac{1}{3}\int_{0}^{t}X_{s}^{\frac{1}{3}}\text{d}s+\int_{0}^{t}X_{s}^{\frac{2}{3}}\text{d}W_{s} $$ However in your case, it is rather easy to see that $X_{0} = 1$, since you are given $$ X_{0} = \left( 1 + \frac{W_{0}}{3} \right)^{3} = 1 $$ since $W_{0}=0$ by properties of the Wiener process.