If $\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$ prove that $\cos\alpha=\frac{2-m^2}{m}$ 
If $$\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$$
  prove that
  $$\cos\alpha=\frac{2-m^2}{m}$$  

My approach:
$$\cos^2(\alpha-3\theta)+\sin^2(\alpha-3\theta)=m^2(\sin^6\theta+\cos^6\theta)$$
$$\Rightarrow \frac{1}{m^2}=\sin^6\theta+cos^6\theta=1-\frac{3}{4}\sin^22\theta$$
$$\Rightarrow \sin^22\theta=\frac{4}{3}᛫\frac{m^2-1}{m}$$
I can not proceed further, please help.

Hint for your last equation:
$$\sin(2\theta)=2sin(\theta)\cos(\theta)$$
so $$\sin(2\theta)^2=4\sin^2(\theta)\cos^2(\theta)$$
and you can eliminate $\theta$
So you get $$4\left(1-\cos(\theta)^2\right)\cos(\theta)^2=\frac{4}{3}\frac{m^2-1}{m}$$
Solve this for $\cos(\theta)$
I know this, when you get $\theta$ then you can compute $\alpha$ with the equations above!
Expanding the term
$$\cos(\alpha-3\theta)\sin(\theta)^3-\sin(\alpha-3\theta)\cos(\theta)^3$$
we get
$$4\, \left( \sin \left( \theta \right)  \right) ^{3}\cos \left( \alpha
 \right)  \left( \cos \left( \theta \right)  \right) ^{3}-3\, \left( 
\sin \left( \theta \right)  \right) ^{3}\cos \left( \alpha \right) 
\cos \left( \theta \right) +4\, \left( \sin \left( \theta \right) 
 \right) ^{4}\sin \left( \alpha \right)  \left( \cos \left( \theta
 \right)  \right) ^{2}- \left( \sin \left( \theta \right)  \right) ^{4
}\sin \left( \alpha \right) -4\, \left( \cos \left( \theta \right) 
 \right) ^{6}\sin \left( \alpha \right) +3\, \left( \cos \left( \theta
 \right)  \right) ^{4}\sin \left( \alpha \right) +4\, \left( \cos
 \left( \theta \right)  \right) ^{5}\cos \left( \alpha \right) \sin
 \left( \theta \right) - \left( \cos \left( \theta \right)  \right) ^{
3}\cos \left( \alpha \right) \sin \left( \theta \right)=0 
$$
This term can we solve for $$\alpha$$
$$\alpha=-\arctan \left( 3\,{\frac {\cos \left( \theta \right) \sin \left( 
\theta \right)  \left( 2\, \left( \sin \left( \theta \right)  \right) 
^{2}-1 \right) }{6\, \left( \sin \left( \theta \right)  \right) ^{4}-6
\, \left( \sin \left( \theta \right)  \right) ^{2}+1}} \right) 
$$ and now you can use your $$\theta$$
