Extreme values of ${x^3 + y^3 + z^3 - 3xyz}$ subject to ${ax + by + cz =1}$ using Lagrange Multipliers
Here I shall find all critical points without verifying what kind of critical points they are (which seems to be your question). My notations are similar to yours, but a bit different, so I shall work from scratch but my answer borrows a lot of your ideas (great attempt, by the way). However, I do not expect that any optimizing point will be a global one. Therefore, be careful and do not assume that any optimizing point will yield a global optimum.
For $X,Y,Z\in\mathbb{R}$, let $$\begin{align}F(X,Y,Z)&:=X^3+Y^3+Z^3-3XYZ\\&=(X+Y+Z)(X^2+Y^2+Z^2-YZ-ZX-XY)\end{align}$$ and $$G(X,Y,Z):=aX+bY+cZ-1\,,$$ where $a$, $b$, and $c$ are fixed real numbers. The task is as follows: $$\begin{array}{ll}\text{optimize}&F(X,Y,Z) \\\text{subject to} & X,Y,Z\in\mathbb{R}\\&G(X,Y,Z)=0\,.\end{array}$$
We set up the Lagrangian $\mathcal{L}(X,Y,Z,\Lambda)$ for $X,Y,Z,\Lambda\in\mathbb{R}$ by $$\mathcal{L}(X,Y,Z,\Lambda):=F(X,Y,Z)-3\,\Lambda\,G(X,Y,Z)\,.$$ If $(x,y,z)\in\mathbb{R}^3$ is such that $(X,Y,Z):=(x,y,z)$ is a solution to this optimization problem, then there exists $\lambda\in\mathbb{R}$ for which $$\frac{\partial \mathcal{L}}{\partial V}(x,y,z,\lambda)=0$$ for all variables $V\in\{X,Y,Z,\Lambda\}$. That is, we have the following equations: $$x^2-yz=\lambda\,a\,,\tag{1}$$ $$y^2-zx=\lambda\,b\,,\tag{2}$$ $$z^2-xy=\lambda\,c\,,\tag{3}$$ and $$ax+by+cz=1\,.\tag{4}$$
We let $f:=F(x,y,z)$. Adding $x$ times (1), $y$ times (2), and $z$ times (3) yields $$\begin{align}f&=x(x^2-yz)+y(y^2-zx)+z(z^2-xy)=x(\lambda\,a)+y(\lambda\,b)+z(\lambda\,c)\\&=\lambda\,(ax+by+cz)=\lambda\cdot 1=\lambda\,,\end{align}$$ due to (4). By adding (1), (2), and (3), we obtain $$\begin{align} f&=(x+y+z)(x^2+y^2+z^2-yz-zx-xy) \\&=(x+y+z)\big((x^2-yz)+(y^2-zx)+(z^2-xy)\big) \\&=(x+y+z)\big(\lambda\,a+\lambda\,b+\lambda\,c)=(x+y+z)(a+b+c)\lambda\\&=(x+y+z)(a+b+c)f\,. \end{align}$$ This means $(x+y+z)(a+b+c)=1$ or $f=0$.
Case I: $a+b+c=0$. Then, $f=0$ must hold (whence $\lambda=f=0$). By adding (1), (2), and (3) together, we obtain $$\frac{(y-z)^2+(z-x)^2+(x-y)^2}{2}=(x^2-yz)+(y^2-zx)+(z^2-xy)=0\,.$$ Thus, $x=y=z$ must be the case. Ergo, $$1=ax+by+cz=ax+bx+cx=(a+b+c)x=0\,,$$ which is a contradiction. Consequently, there does not exist a critical point when $a+b+c=0$.
Case II: $a+b+c\neq 0$ but $f=0$. By adding (1), (2), and (3) together, we conclude, as in Case I, that $x=y=z$. Ergo, $$1=ax+by+cz=ax+bx+cx=(a+b+c)x\text{ implies }x=y=z=\frac{1}{a+b+c}\,.$$ This yields $$(f,x,y,z)=\left(0,\frac1{a+b+c},\frac1{a+b+c},\frac1{a+b+c}\right)\,.\tag{5}$$ We shall see later that, when $a=b=c$, then this is the only case that yields a critical point.
Case III: $a+b+c\neq 0$ and $f\neq 0$. Then, we must have $x+y+z=\dfrac1{a+b+c}$. Subtracting (2) from (1) gives us $(x-y)(x+y+z)=\lambda(a-b)=f(a-b)$, so that $$x-y=(a+b+c)f(a-b)\,.$$ Similarly, $$y-z=(a+b+c)f(b-c)$$ and $$z-x=(a+b+c)f(c-a)\,.$$ Set $k:=(a+b+c)fa-x$. Then, we get $$x=(a+b+c)fa-k\,,\,\,y=(a+b+c)fb-k\,,\text{ and }z=(a+b+c)fc-k\,.$$ As $x+y+z=\dfrac1{a+b+c}$, we obtain $$(a+b+c)^2f-3k=\frac{1}{a+b+c}\,.$$ Because $ax+by+cz=1$, we must have $$(a+b+c)(a^2+b^2+c^2)f-(a+b+c)k=1\,.$$ This shows that $$\begin{align}&(a^3+b^3+c^3-3abc)f\\&\phantom{aaa}=\frac{3\big((a+b+c)(a^2+b^2+c^2)f-(a+b+c)k\big)-(a+b+c)\big((a+b+c)^2f-3k\big)}{2}\\&\phantom{aaa}=\frac{3\cdot 1-(a+b+c)\cdot\left(\frac{1}{a+b+c}\right)}{2}=1\,.\end{align}$$ Consequently, $a^3+b^3+c^3-3abc\neq 0$, implying that $a$, $b$, and $c$ are not all equal, and so $$f=\frac{1}{a^3+b^3+c^3-3abc}\,,\text{ which leads to }k=\frac{bc+ca+ab}{a^3+b^3+c^3-3abc}\,.$$ Thence, $$\begin{align}(f,x,y,z)&=\Biggl(\frac{1}{a^3+b^3+c^3-3abc},\frac{a^2-bc}{a^3+b^3+c^3-3abc}\\&\phantom{aaaaa},\frac{b^2-ca}{a^3+b^3+c^3-3abc},\frac{c^2-ab}{a^3+b^3+c^3-3abc}\Biggr)\,.\end{align}\tag{6}$$