I should determine the spontaneous magnetization of an ferromagnet below its critical temperature $T_{c}$ by only knowing the derivatives:
$$ \left ( \frac{\partial M}{\partial H } \right )_{T}=\frac{a}{1-T/T_c}+3bH^{2} $$ where a and b are some real constants
$$ \left ( \frac{\partial M}{\partial T } \right )_{H}= \frac{1}{T_{c}}\frac{f(H)}{(1-T/T_c)^{2}}- \frac{1}{2}\frac{M_{0}}{T_{c}}\frac{1}{(1-T/T_c)^{1/2}} $$
with $M_{0}$,$T_{c}$,a and b are constants and f(H) is some function with the property f(H=0)=0
a)determine f(H) b)determine M(T,H)
Now I would start by writing down
$$ dM = \left ( \frac{\partial M}{\partial H } \right )_{T}dH + \left ( \frac{\partial M}{\partial T } \right )_{H} dT $$
This tells me how I would obtain the M in terms of the derivatives but I don't know how I would obtain the function f(H). Can anyone give me a hint how I would start the determination of f(H).
$\mathbf{Edit}$: I tried now to do the integration what gives: $$ M(T,H)=\frac{aH}{1-T/T_{c}}+bH^{3}+\frac{f(H)}{1-T/T_{c}}+M_{0}(1-T/T_{c})^{1/2} $$
If I compute now again the derivative with respect to H I obtain: $$ \left ( \frac{\partial M}{\partial H } \right )_{T}=\frac{a}{1-T/T_c}+3bH^{2}+\frac{1}{1-T/T_{c}}\frac{\partial f(H)}{\partial H} $$
This is now not the same as in the exercise sheet. From here I don't know how to go on. Is this already wrong or is it somehow possible from this equation to determine f(H). My first intuition was to say f(H) is just zero, because then the derivative with respect to H would be satisfied and so would be the f(H=0)=0 criterion