= Solution
Neglecting the <electron> <mass> in the centre-of-mass <kinematics>, take $k=(E^*,0,0,E^*)$, $p=(E^*,0,0,-E^*)$, and $k'=(E^*,E^*\sin\theta^\star,0,E^*\cos\theta^\star)$. Then $p\cdot k=2E^{*2}$ and $p\cdot k'=E^{*2}(1+\cos\theta^\star)$. Hence
$$
\boxed{y=1-\frac{p\cdot k'}{p\cdot k}=\frac{1-\cos\theta^\star}{2},\qquad 0\le y\le1.}
$$
The constant <Jacobian determinant>[Jacobian] $|d\cos\theta^\star/dy|=2$ gives
$$
\boxed{\frac{d\sigma_{\mathrm{CC}}}{dy}\propto(1-y)^2,\qquad
\frac1{\sigma_{\mathrm{CC}}}\frac{d\sigma_{\mathrm{CC}}}{dy}=3(1-y)^2.}
$$
Including both Standard Model chiral channels in the <Fermi interaction>[contact approximation] instead gives $d\sigma/dy\propto g_L^2(1-y)^2+g_R^2$. The right-electron term is flat in $y$.
In the electron-rest laboratory frame $p=(m_e,\mathbf0)$, the invariant definition immediately gives
$$
\boxed{y=\frac{m_e(E_\nu-E'_\nu)}{m_eE_\nu}=\frac{E_\nu-E'_\nu}{E_\nu}.}
$$
This last equality is exact. With <electron> <mass> retained in the centre-of-mass relation, the angular formula becomes $y=(s-m_e^2)(1-\cos\theta^\star)/(2s)$, approaching the boxed massless relation at high <energy>.
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