= Solution
A <critical exponent> specifies a leading singular power in a response or <order parameter> as the <thermodynamic critical point> is approached. Analytic backgrounds and the selected thermodynamic branch must be distinguished from that singular part. Write $r=a t$, $a>0$, and first take $u>0$. The equation of state is $h=rM+uM^3$.
At zero field on the ordered branch, $M=(-r/u)^{1/2}$, giving the <order-parameter critical exponent> $\beta=1/2$. Differentiating the equation of state gives $\chi=(r+3uM^2)^{-1}$: above the transition $\chi=1/r$, below it $\chi=1/(2|r|)$, so the <magnetic-susceptibility critical exponent> is $\gamma=1$. At $r=0$, $M=\operatorname{sgn}(h)(|h|/u)^{1/3}$, giving the <critical-isotherm exponent> $\delta=3$.
The minimized singular <free-energy density> is $f_s=0$ above and $f_s=-r^2/(4u)$ below. Two temperature derivatives produce a finite specific-heat jump, corresponding to the <heat-capacity critical exponent> $\alpha=0$. The inverse quadratic fluctuation kernel is $Kp^2+r+3uM^2$, so $\xi=\sqrt{K/(r+3uM^2)}$ gives $\nu=1/2$; the critical propagator proportional to $p^{-2}$ gives $\eta=0$. Thus
$$
\boxed{(\alpha,\beta,\gamma,\delta,\nu,\eta)=(0,\tfrac12,1,3,\tfrac12,0)\quad\text{in the ordinary Landau approximation}.}
$$
These <mean-field critical exponents> follow from minimization and linear response. More generally, linearize the <renormalization group> and differentiate its singular free-energy scaling form to obtain the actual fluctuation-controlled exponents.
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