This is much easier than your comments suggest:
$$\mathbb{E}(M_t^2)=\mathbb{E}(\langle M \rangle_t)=t<\infty$$
Define $X_t = -\int_0^t\beta_s\,dW_s$ and $Y_t = -\frac{1}{2}\int_0^t\beta_s^2\,ds$. Then $Z_t = e^{X_t+Y_t}$. Even though you did not mention it I am guessing that there is a condition on $(\beta_s)_{s\geq 0}$ that makes $X$ a local martingale. Since Brownian motion has continuous paths, $X$ is continuous as well, which implies $Z$ is continuous as well. Hence, (needless to say) $X$ and $Y$ are both semimartingales. Since $(x,y) \mapsto e^{x+y}$ is twice continuously differentiable in all its arguments, Ito's formula applies, i.e.
$$dZ_t = Z_tdX_t + Z_tdY_t + \frac{1}{2}Z_td[X,X]_t+ \frac{1}{2}Z_td[Y,Y]_t + Z_td[X,Y]_t$$
You can check the following identities yourself.
$$dX_t = -\beta_tdW_t$$
$$dY_t = -\frac{1}{2}\beta_t^2dt$$
$$d[X,X]_t = \beta^2_tdt$$
$$d[X,Y]_t = d[Y,Y]_t = 0$$
Substituting these identities into the SDE above,
$$dZ_t = -Z_t\beta_tdW_t - Z_t\frac{1}{2}\beta_t^2dt + Z_t\frac{1}{2}\beta^2_tdt$$
Hence, $dZ_t = -Z_t\beta_tdW_t$. I mentioned above that there must be a condition on $\beta$ that makes $X$ a local martingale. The most general condition for this is that
$$\int_0^t\beta_s^2\,ds < \infty \quad \text{a.s.} \qquad (\triangle)$$
for every $t < \infty$. We will require the same for $Z$ to be a local martingale, i.e.
$$\int_0^tZ_s^2\beta_s^2\,ds < \infty \quad \text{a.s.} \qquad (\square)$$
for every $t < \infty$. Now fix an arbitrary $0 < t < \infty$. Let $\Omega_t$ denote the full measure set where $(\triangle)$ is true. Let $\Omega_c$ be the full measure set where $Z$ is continuous. Fix $\omega \in \Omega_c \cap \Omega_t$. Since $s \mapsto Z_s(\omega)$ is continuous, $s \mapsto 1_{[0,t]}(s)Z_s(\omega)$ is bounded. Then,
$$\int_0^tZ_s(\omega)^2\beta_s(\omega)^2\,ds \leq \sup_{s \in [0,t]}Z_s(\omega)^2 \int_0^t\beta_s(\omega)^2\,ds < \infty$$
Since $\Omega_c \cap \Omega_t$ has full measure, $(\square)$ is true. So then, $Z$ is a local martingale.
Best Answer
I wanted to provide some details on the proof, please let me know any further details are required. To prove that $M_t$ is a local martingale consider the following stopping times: $T_k=\inf\{t: R_t=\frac 1k\}$, $S_k=\inf\{t: R_t=k\}$, $\tau_k=T_k\wedge S_k\wedge n$. We know that $R_t$ satisfies $R_t=\int_{0^t}\frac{d-1}{2R_s}ds + B_t$, where $B_t$ is a Brownian Motion (see Karatzas, Shreve, proposition III.3.21). We can note that stopped processes $B_{t\wedge \tau_k}$ take values within $[1/k,k]$ where function $f$ is twice continuously differentable, thus allowing to apply Ito formula for $f(R_t)$. Doing this, because of the choice of the function $f(x)=x^{2-d}$, we see that $f(R_{t\wedge \tau_k})$ will be a local martingale satisfying $$ M_{t\wedge \tau_k}=M_{1\wedge \tau_k}+(2-d)\int_{1}^{t\wedge \tau_k}R_s^{1-d}dB_s $$ Using the fact that $\tau_k$ is bounded stopping time (because Bessel process of dimension $d\ge 3$ almost surely doesn't reach the origin and $R_t\to \infty$ $(t\to \infty)$) and boundedness of $R_s$ in $s\in (1,\tau_k)$ we conclude that $\int_{1}^{t\wedge \tau_k}R_s^{1-d}dB_s$ is a martingale. Using these arguments we also have that $\tau_k\uparrow \infty$.
Next, we can show that in dimension $d=3$, $M_t=R_t^{-1}$ is not a martingale by computing $\mathbb{E}(M_t)$. $$ \mathbb{E}(M_t)=\int_{\mathbb{R}^3}\frac{1}{\sqrt{x^2+y^2+z^2}}\frac{1}{(\sqrt{2\pi t})^3}\exp\left(-\frac{x^2+y^2+z^2}{2t}\right)dxdydz $$ Rewriting this integral in spherical coordinates $(x,y,z)\mapsto (r, \phi, \theta)$, we can rewrite it as $$ \frac{1}{(\sqrt{2\pi t})^3}\int_{0}^\infty\int_{0}^{2\pi}\int_{0}^{\pi}\frac 1r\exp(-\frac{r^2}{2t})r^2\sin\theta drd\phi d\theta= $$ $$ =\frac{1}{(\sqrt{2\pi t})^3}\int_{0}^\infty\frac 1r\exp(-\frac{r^2}{2t})r^2 dr\cdot 2\pi \int_{0}^{\pi}\sin \theta d\theta= $$ $$ =\frac{2\cdot 2\pi t}{(\sqrt{2\pi t})^3}\int_{0}^\infty\exp(-u) du =\sqrt{\frac{2}{\pi t}} $$ If $M_t$ were martingale then $\mathbb{E}(M_t)$ had to be the same for all $t\ge 1$, but above computation show that this expectation depends on $t$.