= Solution
Differentiating the <grand potential> gives
$$
N=kVT^bf'(y),\qquad S=kVT^b[(b+1)f(y)-yf'(y)],\qquad
\frac SN=(b+1)\frac{f(y)}{f'(y)}-y.
$$
A reversible adiabatic change in an isolated gas keeps $S$ and $N$ fixed. To infer constant $y$, one must check that this ratio is one-to-one, rather than merely assuming it.
Put $q=y/k$, $n(x)=1/(1+e^{x-q})$, $F_j=\int_0^\infty x^jn(x)\,dx$ and $M_j=\int_0^\infty x^{a+j}n(x)(1-n(x))\,dx$ for $j=0,1,2$. <Integration by parts> gives $M_1=bF_a$ and $M_2=(b+1)F_{a+1}$, and
$$
\frac S{kN}=\frac{b+1}{b}\frac{F_{a+1}}{F_a}-q,\qquad
\frac d{dq}\frac S{kN}=b\left(1-\frac{M_0M_2}{M_1^2}\right)<0.
$$
The strict inequality is the <Cauchy-Schwarz inequality> for the positive weight $x^an(x)(1-n(x))$, since $x$ is not constant on its support. This proves <strict monotonicity of Fermi gas entropy per particle> and hence $\mu/T$ is constant. Then $N=kVT^bf'(y)$ gives $VT^b$ constant, and $P=kT^{b+1}f(y)$ gives
$$
\boxed{VT^c=\text{constant},\quad PV^d=\text{constant},\qquad c=a+1,\quad d=\frac{a+2}{a+1}.}
$$
This <adiabatic scaling of a Fermi gas with power-law density of states> concerns positive temperature; the zero-temperature <limit> is a separate limiting case.
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