Finding critical values of a function on an embedded surface Prior to the problem, we have already shown that $\Sigma=\{x_1x_2^2+x_2x_3^2+x_3x_1^2=1\}\subset\mathbb{R}^3$ is an embedded hypersurface and that the function $f:\Sigma\rightarrow\mathbb{R};(x_1,x_2,x_3)\mapsto x_1$ is surjective. The problem asks to find the critical values of $f$.
If we had a "nice" global coordinate chart on $\Sigma$, which is 2-dimensional, we could calculate the differential of $f$ in these coordinates and find out where it vanishes. The canonical coordinates inherited from $\mathbb{R}^3$ are however 3-dimensional and not appropriate for this purpose. Therefore, we tried to use the following trick:
Define $F:\mathbb{R}^3\rightarrow\mathbb{R};F(x_1,x_2,x_3)=x_1x_2^2+x_2x_3^2+x_3x_1^2$ and $E_a:=\{(a,x_2,x_3)\in\mathbb{R}^3|x_2,x_3\in\mathbb{R}\}$ for some $a\in\mathbb{R}$. We claim that the critical points of $f$ are exactly the critical points of $F|_{E_a}$ which also lie in $\Sigma$.
Assuming this claim to be true, we calculated
\begin{equation}
d(F|_{E_a})_{(x_2,x_3)}=\begin{pmatrix}x_3^2+2ax_2&&2x_2x_3+a^2\end{pmatrix}
\end{equation}
If $a=0$, then $d(F|_{E_a})_{(x_2,x_3)}=0$ implies $x_3=0$ and $x_2$ is arbitrary, but this triple does not lie on $\Sigma$.
If $a\neq 0$, $d(F|_{E_a})_{(x_2,x_3)}=0$ implies $x_3=a$ and $x_2=-a/2$, then $(a,-a/2,a)\in\Sigma$ implies $a=\sqrt[3]4$. Hence, $\sqrt[3]4$ is the only critical value of $f$.
The question is: Is the claim correct? If yes, how can we prove it? If no, how can we solve the original problem?
 A: This is indeed a Lagrange multipliers problem, but I'd rather describe it as follows. The critical points $p\in\Sigma=\{F=1\}$ of $\ f|\Sigma\ $ are given by $d_p(f|\Sigma)\equiv0$. Since $d_p(f|\Sigma)=d_af|T_a\Sigma$, that $\equiv0$ exactly means $d_pf|T_p\Sigma\equiv0$. As $T_p\Sigma=\ker(d_pF)$ we conclude that the critical points $p$ are given by the condition
$$
\ker(d_pf)\supset\ker(d_pF).
$$
Now, the LHS kernel is the orthogonal complement of $\nabla f$, and the RHS kernel is that of $\nabla F$. Thus the condition is equivalent to that
$\nabla f$ is a multiple of $\nabla F$ (this is Lagrange multipliers). This can be written
$$
\text{rk}\begin{pmatrix}
\frac{\partial f}{\partial x_1}&\frac{\partial f}{\partial x_2}&\frac{\partial f}{\partial x_3}\\
\frac{\partial F}{\partial x_1}&\frac{\partial F}{\partial x_2}&\frac{\partial F}{\partial x_3}
\end{pmatrix}=1
$$
Thus
$$
\text{rk}\begin{pmatrix}
1&0&0\\
x_2^2+2x_3x_1&x_3^2+2x_2x_1&x_1^2+2x_3x_2
\end{pmatrix}=1,
$$
which means $x_3^2+2x_2x_1=x_1^2+2x_3x_2=0$. This is your claim!
Now the point is why this is your claim. In fact, what you have computed are the critical points $p=(a,x_2,x_3)\in E_a$ of $F|E_a$, where $E_a=\{f=a\}$. Since $F|E_a$ is just a two variables function $F(a,x_2,x_3)$ you get the result straight away. But if we use the general argument above, these critical points are given by
$$
\ker(d_pf)\subset\ker(d_pF).
$$
Thus we have reversed the inclusion, which is then equivalent to that
$\nabla F$ is a multiple of $\nabla f$. In our case, since both gradients are non-zero, it doesn't matter which is multiple of which and we understand why teh critical points are the same in both cases. 
Now this is a particular situation. If, for instance, $\Sigma$ was a curve defined by a mapping $F=(F_1,F_2):\mathbb R^3\to\mathbb R^2$, then the critical points
of $f|\Sigma$ would be given by the condition that $\nabla f$ was a linear combination $\sum_i\lambda_i\nabla F_i$ of the gradients $\nabla F_i$: these $\lambda_i$ are the famous Lagrange multipliers. 
