Consider the family: $$ \begin{matrix} X & = & \textbf{Proj}\left(\frac{\mathbb{C}[s,t,u][x,y,z]}{s(x^4 -y^2z^2) + t(x^2y^2 - z^4) - u(x^4 + y^4 + z^4)}\right) & \\ \downarrow & & \downarrow\\ B & = & \textbf{Proj}(\mathbb{C}[s,t,u]) \end{matrix} $$ How can I find the hyperplane $H_x \subset B$ of points whose fibers contain a point $x \in X$? For example, let $x = [0:1:1:1:(-1)^{(1/3)}:(-1)^{(1/3)}]$. For reference, I am looking at Nicolaescu's introduction to morse theory (page 252).
For reference, here is what I have tried: if I have the point $[s:t:u:a:b:c] \in X$, then I think the hyperplane will be spanned by the set of points $$ \{s,t,u \in \mathbb{C}: s(a^4 - b^2c^2) + t(a^2b^2 - c^4) + u(a^4 + b^4 + c^4)\} $$ I am guessing this because of the statement afterwards $$ H_x = \{ P \in U : P(x) = 0 \} $$ where $U$ is the projective plane parametrizing this family, but I am not sure why this is true.
If the fiber over $[s:t:u]\in B$ contains $x=[s':t':u':a:b:c]\in X$ then this means that, denoting by $P$ the equation of $X$ you wrote :
$$[s:t:u]=[s':t':u']\text{ and }P(s',t',u',a,b,c)=0.$$
In fact, by definition of $X\to B$, the fiber over $[s_0:t_0:u_0]\in B$ is $$\{[x:y:z]\in \Bbb P^2(\Bbb C) : P(s_0,t_0,u_0,x,y,z)=0\}\subseteq \mathrm{Proj}(\Bbb C[s_0,t_0,u_0][x,y,z])\subseteq \mathrm{Proj}(\Bbb C[s,t,u][x,y,z]).$$
So, what you wrote is almost correct : I would write $H_x=\{ [s:t:u]\in \Bbb P^2(\Bbb C) : s(a^4 - b^2c^2) + t(a^2b^2 - c^4) + u(a^4 + b^4 + c^4)\}$ instead of $\{ s,t,u\in\Bbb C : \dots\}$.
Edit: Now, denoting by $\hat{U}$ the dual projective space of $U$, if we want to describe explicitly the map $X\to \hat{U}$ defined by $x\mapsto H_x$, we can write it in coordinates. For this purpose, note that if $(As+Bt+Cu=0)$ is the equation of a hyperplane $H$ of $\text{Proj}(\Bbb C[s,t,u])$, then the coordinates of $H$ in $\hat{U}$ are $[A:B:C]$.
Hence, in your example, the map $x\mapsto H_x$ is defined by $$[s:t:u:a:b:c]\mapsto [a^4 - b^2c^2:a^2b^2 - c^4:a^4 + b^4 + c^4].$$