= Solution
The <scaling hypothesis for critical phenomena> says that the singular part of the equilibrium <free energy> is a generalized homogeneous function of its thermal and field-like controls. Smooth backgrounds must first be removed; the hypothesis does not claim that the entire <free energy>, including arbitrary regular terms, has a pure scaling form.
For the ordinary scalar quartic <LG theory>, set $r=r_t t$ with $r_t>0$ and $u>0$, and rescale the uniform <order parameter> as
$$
m=\sqrt{\frac{r_t}{u}}\,|t|^{1/2}\psi.
$$
The order-parameter-dependent <free-energy density> becomes
$$
V(m)=\frac{r_t^2}{u}|t|^2\left\{\frac{\operatorname{sgn}(t)}2\psi^2+\frac14\psi^4-H\psi\right\},
\qquad H=\frac{\sqrt u}{r_t^{3/2}}\frac h{|t|^{3/2}}.
$$
Define $f_\pm(H)$ as the minimum of the braces, with the sign of the quadratic term respectively positive or negative. Consequently
$$
\boxed{A_s=a|t|^2f_\pm\left(b\frac h{|t|^{3/2}}\right),\quad a=\frac{r_t^2}{u}>0,\quad b=\frac{\sqrt u}{r_t^{3/2}}>0.}
$$
For an extensive $A_s$, $a$ additionally contains the system volume. The printed double-inequality subscript labels the two temperature branches: the upper-temperature function is $f_+$ and the lower-temperature function is $f_-$. It does not classify positive and negative magnetic fields. In particular $f_+(0)=0$ and $f_-(0)=-1/4$; below the transition the field dependence has a cusp at zero, so derivatives are taken on a selected branch.
For the following derivatives, take $A_s$ to be a density, so $m$ is the <magnetization> density; for total <free energy> the derivative gives total <magnetization> instead. Differentiate this <mean-field scalar free-energy scaling> form. The <magnetization> is $m=-\partial A_s/\partial h=-ab|t|^{1/2}f_\pm'(H)$, giving $m_0\propto(-t)^{1/2}$ and \b[$\beta=1/2$]. The <magnetic susceptibility> is $\chi=-\partial_h^2A_s=-ab^2|t|^{-1}f_\pm''(H)$, giving \b[$\gamma=1$]. At zero field the nonzero curvature amplitudes are $-f_+''(0)=1$ and $-f_-''(0^+)=1/2$.
To allow nonclassical <critical exponents>, replace the fixed powers by
$$
\boxed{A_s=a|t|^{2-\alpha}f_\pm\left(b\frac h{|t|^{\Delta}}\right).}
$$
The <heat-capacity critical exponent> is defined by $C_{V,s}\sim|t|^{-\alpha}$; temperature differentiation gives the thermal exponent $2-\alpha$ in $A_s$. The <order-parameter critical exponent> has $m_0\sim(-t)^\beta$, and the <magnetic-susceptibility critical exponent> has $\chi\sim|t|^{-\gamma}$. Differentiating the scaling form gives
$$
\beta=2-\alpha-\Delta,\qquad\gamma=2\Delta-(2-\alpha).
$$
Eliminating $\Delta$ proves the <Rushbrooke scaling relation>
$$
\boxed{\alpha+2\beta+\gamma=2.}
$$
The <critical-isotherm exponent> is defined by $m(0,h)\sim\operatorname{sgn}(h)|h|^{1/\delta}$. At fixed small $h$, the $t\to0$ limit requires $f_\pm(H)\sim c|H|^{(2-\alpha)/\Delta}$, so that the temperature factors cancel. Therefore $1/\delta=(2-\alpha-\Delta)/\Delta=\beta/\Delta$, yielding $\Delta=\beta\delta$. But the two differentiated identities also give $\Delta=\beta+\gamma$, and hence the <Widom scaling relation>
$$
\boxed{\beta\delta=\beta+\gamma.}
$$
These are relations among the leading power indices. At marginal dimensions, multiplicative logarithms can accompany them, and an additive analytic background must not be mistaken for the singular scaling contribution.
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