The question: Use substitution $z=\frac{1}{y^3}$ to reduce the equation:
$$\sec\left(x\right)\:\frac{dy}{dx}+y=y^4$$
to a linear form.
I'm stuck on whether to first divide by some function of $y$ then substitute, and how to rewrite $\frac{dy}{dx}$ in terms of $z$.
For the substitution $z:=\frac{1}{y^3}$ (which is a substitution inspired by the ode which is Bernoulli) with chain rule one can write $$\frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}=(-\frac{3}{y^4}) \frac{dy}{dx}\leadsto \frac{dy}{dx}=(-\frac{y^4}{3})\frac{dz}{dx}.$$
Yes, you "first divide for $y^4$" in the ode to obtain $$\frac{1}{y^4}\sec(x)y'+\frac{1}{y^4}y=\frac{y^4}{y^4},\quad y\not=0$$ Then return to the substitution and one can write $$\frac{1}{y^4}\sec(x)(-\frac{y^4}{3}\frac{dz}{dx})+\frac{1}{y^4}y=1\leadsto-\frac{\sec x}{3}\frac{dz}{dx}+z=1$$ Then you can re-write it as $$\frac{dz}{dx}-3\cos(x)z=\cos(x),\quad \cos(x)\not=0$$ which is linear with general solution $$z=1+Ae^{3\sin x}\leadsto \frac{1}{y^3}=1+Ae^{3\sin x}$$
NB: I suppose that the problem is to work on the substitution, in any case it does not hurt to comment that your ode is separable from the beginning $$\frac{dy}{dx}=\cos(x)(y^4-y)$$ Then integrating give the answer.