= Solution
For the stationary target, $p=(m_N,\mathbf0)$. The invariant definition immediately gives
$$
\boxed{y=\frac{p\cdot(k-k')}{p\cdot k}
=\frac{m_N(E_\nu-E_\mu)}{m_NE_\nu}
=1-\frac{E_\mu}{E_\nu}.}
$$
No <massless particle>[massless] approximation is needed for this laboratory identity when binding and target recoil conventions are neglected.
For the angular relation, specify the neutrino-parton <centre-of-momentum frame> and neglect <masses> there. The incoming and outgoing <leptons> then each have <energy> $\sqrt{\widehat s}/2$. Their <momentum> transfer satisfies
$$
Q^2=2\left(\frac{\sqrt{\widehat s}}2\right)^2(1-\cos\theta^\star)
=\frac{\widehat s}{2}(1-\cos\theta^\star).
$$
Since $Q^2=xys$ and $\widehat s=xs$, cancellation gives
$$
\boxed{y=\frac{1-\cos\theta^\star}{2},\qquad
1-y=\frac{1+\cos\theta^\star}{2}.}
$$
This identifies the <antiquark> weight $(1-y)^2$ with the corresponding squared angular factor. The relevant <massless particle>[massless] condition is sufficient <energy> in the lepton-parton subprocess; $E_\nu\gg m_\mu$ alone does not control every endpoint configuration.
The frame qualifier matters: the complete neutrino-nucleon <centre-of-momentum frame>[centre-of-mass] final state contains a variable-mass <hadron>[hadronic] system. In that frame, still neglecting <masses> but retaining $x$, direct two-body kinematics for $\mu+X$ gives instead
$$
y=\frac{1-\cos\theta_{\nu N}}{1-\cos\theta_{\nu N}+x(1+\cos\theta_{\nu N})}.
$$
For example $x=1/2$ and $\theta_{\nu N}=\pi/2$ give $y=2/3$, not $1/2$. Thus the simple angle-only relation implicitly uses the <parton model>[partonic] frame.
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