= Solution
In the <parton model>, the incoming parton has $k=\xi P_H$ and the outgoing massless parton has $k'=\xi P_H+q$. Neglecting $P_H^2$ and imposing its on-shell condition gives
$$
0=(\xi P_H+q)^2=2\xi P_H\cdot q+q^2,
\qquad \boxed{\xi=\frac{-q^2}{2P_H\cdot q}=x.}
$$
Thus the <Bjorken scaling variable> measures the struck parton's longitudinal momentum fraction in this approximation. The <inelasticity> is unchanged when the hadron momentum is replaced by its parton's collinear momentum:
$$
\boxed{y_q=\frac{\xi P_H\cdot q}{\xi P_H\cdot p}=\frac{P_H\cdot q}{P_H\cdot p}=y.}
$$
Let $S_H=(p+P_H)^2\simeq2p\cdot P_H$. The partonic invariant in part (b) is $s(\xi)=\xi S_H$, and $Q^2=-q^2=xyS_H$. Insert its two parton cross sections into the <parton distribution function> convolution:
$$
\frac{d\sigma_H}{dy}=\frac{G_F^2S_H}{\pi}\int_0^1\xi\left[q_d(\xi)+(1-y)^2q_{\bar u}(\xi)\right]d\xi.
$$
Since the event's measured $x$ fixes $\xi=x$, the integrand is the cross-section density in $x$. No additional Jacobian from $y_q$ is needed because $y_q=y$. Therefore
$$
\boxed{\frac{d^2\sigma_H}{dy\,dx}=\frac{G_F^2S_H}{\pi}\,x\left[q_d(x)+(1-y)^2q_{\bar u}(x)\right].}
$$
Equivalently $S_H=2E_\nu E_H(1-\cos\vartheta_{\nu H})$ in the chosen massless-hadron frame. The explicit factor $x$ comes from the partonic energy $s=xS_H$, not from redefining the <parton distribution functions>. This expression uses the stated two-flavor and low-energy approximations, with masses and generation mixing omitted.
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