Here is the $2$-descent, following Silverman, The Arithmetic of Elliptic Curves, Proposition X.4.9 and Example X.4.10.
Let $E: y^2 = x^3 + 6x^2 + x$ and $E': Y^2= X^3-12X^2+32X$ and $\phi: E \to E', (x,y) \mapsto(y^2/x^2,y(1-x^2)/x^2)$. This has the $\mathbf{Q}$-rational $2$-torsion point $(0,0)$. Then in Silverman's notation, one has $a = 6, b = 1$. The discriminant is $\Delta_E = 2^9$, so $S = \{2,\infty\}$ and $\mathbf{Q}(2,S) = \{\pm1,\pm2\}$.
For $d \in \mathbf{Q}^\times$, one has the principal homogeneous space $C_d: dw^2 = d^2- 12dz^2+32z^4$. The Selmer group is $\mathrm{Sel}^{(\phi)}(E/\mathbf{Q}) = \{d \in \mathbf{Q}(2,S):C_d(\mathbf{Q}_v) \neq \emptyset\,\,\forall v \in S\}$.
If $d < 0$, clearly $C_d(\mathbf{R}) = \emptyset$. (For $d = 1$, $C_1: w^2 = 1 - 12z^2 + 32z^4$ has the $\mathbf{Q}$-rational point $(w,z) = (0,\frac{1}{2})$, so $1 \in \mathrm{Sel}^{(\phi)}(E/\mathbf{Q})$.) For $d = 2$, $C_1: 2w^2 = 4 - 24z^2 + 32z^4$ has the $\mathbf{Q}$-rational point $(w,z) = (0,\frac{1}{2})$, so $2 \in \mathrm{Sel}^{(\phi)}(E/\mathbf{Q})$.
Hence $\mathrm{Sel}^{(\phi)}(E/\mathbf{Q}) \cong \mathbf{Z}/2$. One has the exact sequence $0 \to E'(\mathbf{Q})/\phi(E(\mathbf{Q})) \to \mathrm{Sel}^{(\phi)}(E/\mathbf{Q}) \to \mathrm{III}(E/\mathbf{Q})[\phi] \to 0$. $E'$ has the $\mathbf{Q}$-rational $2$-torsion point $(0,0)$, so $\mathrm{III}(E/\mathbf{Q})[\phi] = 0$ and $E'(\mathbf{Q})/\phi(E(\mathbf{Q})) \cong \mathbf{Z}/2$ generated by $(0,0)$. Similarly, $\mathrm{III}(E'/\mathbf{Q})[\hat{\phi}] = 0$. One gets $E(\mathbf{Q})/2 \cong \mathbf{Z}/2$ and that the rank is $0$ and the Mordell-Weil groups are $E(\mathbf{Q}) \cong \mathbf{Z}/4$ and $E'(\mathbf{Q}) \cong (\mathbf{Z}/2)^2$.
(In fact, Magma calculations show that the Tate-Shafarevich group is trivial.)