Let $\{W_t\}_{t \ge 0}$ be a standard Brownian motion on a probability space $(\Omega,\cal{F},\Bbb P)$ and let $\{\cal F_t\}_{t \ge 0}$ be the filtration of the Brownian motion. Use the Ito-Doeblin Lemma to derive the dynamic of the stochastic process $Y_t := W_t^3, t \ge 0$.
Attempt: The Ito-Doeblin Lemma is given by $$df(W_t)=f'(W_t)dW_t+\frac12f''(W_t)dt, \qquad \qquad (1)$$ where $f(x)$ is a differentiable function for which the derivatives $f'$ and $f''$ are defined and continuous.
Given $Y_t:=W_t^3, t \ge 0$. Let $f(x)=x^3$. Then $f'(x)=3x^2$ and $f''(x)=6x$. Hence, by $(1)$, we obtain $$dY_t=df(W_t)=3W_t^2dW_t+\frac12 \cdot 6W_tdt = 3W_t^2dW_t + 3W_tdt.$$ Therefore, the dynamic of the stochastic process $Y_t=W_t^3, t \ge 0$ is given by $$dY_t=\alpha_t dt + \sigma_tdW_t,$$ where $\alpha_t=3W_t^2$ and $\sigma_t=3W_t$.
Am I true?