Considering $s$ is implicit to function of $p$, given by $s^6 - p^4 = 1$. For what $s$ is it increasing and decreasing?
Well, I answered first like following: Calculating the first derivative using the implicit differentiation is $s'=\large\frac{2p^3}{3s^5}$. $s$ is increasing where $s' > 0$ and decreasing where $s' < 0$.
Obviously, I answered too generally and I'm wondering how I better answer it using the first derivative to support my answer.
We are talking here about the solution set $$S:=\bigl\{(p,s)\>\bigm|\>F(s,p)=0\bigr\}\ ,$$ where $$F(p,s)=s^6-p^4-1\ .$$ The implicit function theorem says the following: When $(p_0,s_0)\in S$ and the "technical condition" $${\partial F\over\partial s}\biggr|_{(p_0,s_0)}=6s_0^5\ne0\tag{1}$$ is satisfied then there is a window $$W:=\ ]p_0-h,p_0+h[\ \times\ ]s_0-h', s_0+h'[\ $$ and a $C^1$-function $$\psi:\quad ]p_0-h,p_0+h[\ \to\ ]s_0-h', s_0+h'[\ ,$$ such that $S\cap W$ equals the graph of $\psi$. Furthermore one has $$\psi'(p_0)=-{F_p(p_0,s_0)\over F_s(p_0,s_0)}={2p_0^3\over 3s_0^5}\ .$$ Now $S$ contains no points with $s=0$; therefore $(1)$ is satisfied at all points of $S$.
Whether the local function $\psi$ is increasing or decreasing in its window depends on the signs of $p_0$ and $s_0$: When $p_0$ and $s_0$ have the same sign $\psi$ is increasing (for small enough $h>0$), otherwise $\psi$ is decreasing. The points $(0,\pm1)\in S$ are special, since $\psi'(0)=0$ there. Here further analysis is necessary, which I leave to you.