Yes, the process $(X_t)$ defined by $X_t=\displaystyle\int_0^tB_s^2\mathrm dB_s$ is a martingale, as every square integrable stochastic integral $\displaystyle\int_0^tu(s,B_s)\mathrm dB_s$, but, no, going back to Itô's formula to show this is not useful since this is going backwards because one already knows that $\mathrm dX=B^2\mathrm dB$, which is the kind of conclusion Itô's formula can help to reach.
No, the process $(Y_t)$ defined by $Y_t=B_t^3$ is not a martingale since, for every $t\gt s\gt0$, the decomposition $B_t=B_s+(B_t-B_s)$ with $B_s$ measurable with respect to $\mathcal F_s$ and $B_t-B_s$ independent of $\mathcal F_s$ yields the almost sure identity $E(B_t^3\mid\mathcal F^B_s)=B_s^3+3(t-s)B_s\ne B_s^3$.
No, a process being a supermartingale does not imply that it is a martingale, and no, the process $-\frac13B^3$ is not a supermartingale (and not a submartingale either).
No, it's not exactly obvious. Let's prove the following theorem.
Let $(B_t)_{t \geq 0}$ be a one-dimensional Brownian motion and $(K_t)_{t \geq 0}$ a progressively measurable process such that $\mathbb{E}\int_0^t K_s^2 \, ds < \infty$ for all $t \geq 0$. If $(N_t)_{t \geq 0}$ is a continuous $L^2$-bounded martingale, then $$M_t := \int_0^t K_s \, dB_s$$ satisfies $$\langle M,N \rangle_t = \int_0^t K_s \, d\langle B,N \rangle_s.$$
Proof: Througout, $(N_t)_{t \geq 0}$ is an $L^2$-bounded martingale. First we consider the particular case that $K$ is a simple process of the form $$K(s) = \sum_{j=0}^{N-1} \varphi_j 1_{(s_j,s_{j+1}]}(s) \tag{1}$$ where $0<s_0 < \ldots < s_N$ and $\varphi_j \in L^2(\mathcal{F}_{s_j})$. We have to show that $M_t = \int_0^t K_r \, dB_r$ satisfies $$\mathbb{E}((M_t-M_s) (N_t-N_s) \mid \mathcal{F}_s) = \mathbb{E} \left( \int_s^t K_r \, d\langle B,N \rangle_r \mid \mathcal{F}_s \right) \tag{2}$$ for any fixed $s \leq t$. Without loss of generality, we may assume that $s_N = t$ and that there exists $k \in \{0,\ldots,N\}$ such that $s_k = s$ (otherwise we refine the partition accordingly). Writing $$N_t-N_s = \sum_{i=k}^{N-1} (N_{s_{i+1}}-N_{s_i}) \quad \text{and} \quad M_t-M_s = \sum_{j=k}^{N-1} (M_{s_{j+1}}-M_{s_j})$$ we find
$$\mathbb{E}((M_t-M_s) (N_t-N_s) \mid \mathcal{F}_s) = \sum_{j=k}^{N-1} \sum_{i=k}^{N-1} \mathbb{E}((M_{s_{j+1}}-M_{s_j}) (N_{s_{i+1}}-N_{s_i}) \mid \mathcal{F}_s).$$
Since both $(M_t)_{t \geq 0}$ and $(N_t)_{t \geq 0}$ are martingales, it is not difficult to see from the tower property that the terms on the right-hand side vanish for $i \neq j$, and so
$$\mathbb{E}((M_t-M_s) (N_t-N_s) \mid \mathcal{F}_s) = \sum_{j=k}^{N-1} \mathbb{E}(\varphi_j (B_{s_{j+1}}-B_{s_j}) (N_{s_{j+1}}-N_{s_j}) \mid \mathcal{F}_s).$$
Using once more the tower property, we get
$$\begin{align*} \mathbb{E}((M_t-M_s) (N_t-N_s) \mid \mathcal{F}_s) &= \sum_{j=k}^{N-1} \mathbb{E} \bigg[ \varphi_j \mathbb{E}((B_{s_{j+1}}-B_{s_j}) (N_{s_{j+1}}-N_{s_j}) \mid \mathcal{F}_{s_j}) \mid \mathcal{F}_s \bigg] \\ &= \sum_{j=k}^{N-1} \mathbb{E} \bigg[ \varphi_j \mathbb{E}(\langle B,N \rangle_{s_{j+1}}-\langle B,N \rangle_{s_j}) \mid \mathcal{F}_{s_j}) \mid \mathcal{F}_s \bigg]\\ &= \mathbb{E} \left( \sum_{j=0}^{N-1} \varphi_j (\langle B,N \rangle_{s_{j+1}}-\langle B,N \rangle_{s_j}) \mid \mathcal{F}_s \right) \\ &= \mathbb{E} \left( \int_s^t K_r \, d\langle B,N \rangle_r \mid \mathcal{F}_s \right) \end{align*}$$
This proves the assertion for the simple process $K$. For the general case we choose a sequence of simple process $(K_n)_{n \in \mathbb{N}}$ of the form $(1)$ such that $$\mathbb{E} \left( \int_0^t (K_n(s)-K(s))^2 \, ds \right) \to 0.$$ By the construction of the stochastic integral, this implies, in particular,
$$ \int_0^t K_n(r) \, dB_r \to \int_0^t K(r) \, dB_r \quad \text{in $L^2(\mathbb{P})$} \tag{3}$$
On the other hand, it follows from the Cauchy-Schwarz inequality that
$$\int_0^t K_n(r) \, d\langle B,N \rangle_r \to \int_0^t K(r) \, d\langle B,N \rangle_r \quad \text{in $L^2(\mathbb{P})$} \tag{4}$$
Thus, $M_t = \int_0^t K(r) \, dB_r$ satisfies
$$\begin{align*} \mathbb{E} \left( (M_t-M_s) (N_t-N_s) \mid \mathcal{F}_s \right) &\stackrel{(3)}{=} \lim_{n \to \infty} \mathbb{E} \left[ \left( \int_0^t K_n(r) \, dB_r - \int_0^s K_n(r) \, dB_r \right) (N_t-N_s) \mid \mathcal{F}_s \right] \\ &\stackrel{(1)}{=} \lim_{n \to \infty} \mathbb{E} \left( \int_s^t K_n(r) \, d\langle B,N \rangle_r \mid \mathcal{F}_s \right) \\ &\stackrel{(4)}{=} \mathbb{E} \left( \int_s^t K(r) \, d\langle B,N \rangle_r \mid \mathcal{F}_s \right). \end{align*}$$
Best Answer
The Lemma is the right one and we need to show the second equals sign in $$\tag{1} \int_0^te^{\alpha_s}\,dB_{\alpha_s}=\int_0^{\alpha_t}e^s\,dB_s=\int_0^tr\,d\bar{B}_r\,. $$
With $f(t):=\alpha'_t$ we define the continuous martingale $$\tag{2} \bar{B}_t:=\int_0^t\frac{1}{\sqrt{f(r)}}\,dB_{\alpha_r} $$ which has quadratic variation $$\tag{3} \langle \bar B\rangle_t=\int_0^t\frac{1}{f(r)}\, d\alpha_r=\int_0^t\frac{1}{\alpha'_r}\alpha'_r\,dr=t $$ and is therefore a Brownian motion. Clearly by (2), $$\tag{4} \int_0^t\frac{r}{\sqrt{f(r)}}\,dB_{\alpha_r}=\int_0^tr\,d{\bar B}_r\,. $$ Finally, $$ \alpha'_t=\frac{t^2}{1+\frac{2}{3}t^3}\,,\quad\frac{1}{\sqrt{f(t)}}=\frac{\sqrt{1+\frac{2}{3}t^3}}{t}=\frac{e^{\alpha_t}}{t}\,,\quad\frac{t}{\sqrt{f(t)}}=e^{\alpha_t}\,. $$ Putting the last expression into (4) shows (1).
To elaborate on (3): This follows from two facts:
The quadratic variation of $\int_0^tY_s\,dM_s$ ($M$ a local martingale) is $\int_0^tY^2_s\,d\langle M\rangle_s\,.$
In the present case $M_t=B_{\alpha_t}$ and the second fact is $\langle B_{\alpha_.}\rangle_t=\alpha_t\,.$ To see why this is true observe that for any continuous increasing function $t\mapsto\alpha_t$
$$ \langle B_{\alpha_.}\rangle_t=\lim_{n\to\infty}\sum_{i=1}^n\big(B_{\alpha_{t_i}}-B_{\alpha_{t_{i-1}}}\big)^2=\alpha_t $$ because, as we know, $$ \langle B\rangle_T=\lim_{n\to\infty}\sum_{i=1}^n\big(B_{T_i}-B_{T_{i-1}}\big)^2=T $$ and $\alpha_{t_0},...,\alpha_{t_n}$ is a sequence of finer and finer partitions of the interval $[0,\alpha_t]\,.$ Just take $T=\alpha_t$ and $T_i=\alpha_{t_i}\,.$