Given a point $P_1=(s_1,t_1)$ on a cubic curve given by $s=t^3+at^2s+bts^2+cs^3$ I need to calculate the slope of the tanget line at $P_1$. This is part of a proof of the Nagell-Lutz Theorem. The book states that the slope equals $$\alpha=\frac{ds}{dt}(P_1)= \frac{3t_1^2+2at_1s_1+bs_1^2}{1-at_1^2-2bt_1s_1-3cs_1^2}$$
But I have no idea how to calculate this derivate, since it seems like I can not solve this given equation for $s$.
I also tried to apply the implicit function theorem. For $$f(t,s)= t^3+at^2s+bts^2+cs^3-s$$ I know $f(t_1,s_1)=0$. Obviously $$D_sf(t,s)=(at^2+2bts+3cs^2-1)$$ but I struggle to verify that $\det D_sf(t_1,s_1)\not=0$. And even if I could I am not sure how I would receive this final form for $\frac{ds}{dt}$, since this theorem does not provide a closed form for $s$ in a implicit area around $P_1$, but only states that there is a closed form. Or am I messing up something?
Any help would be highly appreciated.