Extreme values of ${x^3 + y^3 + z^3 - 3xyz}$ subject to ${ax + by + cz =1}$ using Lagrange Multipliers 
If ${ax + by + cz =1}$, then show that in general ${x^3 + y^3 + z^3 - 3xyz}$ has two stationary values ${0}$ and $\frac{1}{(a^3+b^3+c^3-3abc)}$, of which first is max or min according as ${a+b+c>0}$ or ${< 0}$ but second is not an extreme value. Comment on particular cases when (i) ${a+b+c=0}$, (ii) ${a=b=c}$.

My Attempt:
$${F=f+\lambda \phi =x^3 + y^3 + z^3 - 3xyz + 3\lambda(ax + by + cz-1)}$$
$${\frac13F_x = x^2-yz+\lambda a = 0}{\text{ ...(1)}}$$
$${\frac13F_y = y^2-xz+\lambda b = 0}{\text{ ...(2)}}$$
$${\frac13F_z = z^2-xy+\lambda c = 0{\text{ ...(3)}}}$$
$${(1)x+(2)y+(3)z \implies f+\lambda (1) = 0 \implies \lambda = -f}$$
$${(1)+(2)+(3) \implies}{x^2+y^2+z^2-xy-yz-zx=(a+b+c)f}$$ 
$${\implies f/(x+y+z)=(a+b+c)f}\,,$$ 
then ${f=0}$ or ${x+y+z=1/(a+b+c)}$.
Also, ${(1)-(2) \implies x^2-y^2-z(x-y)=f(a-b) \implies \frac{x-y}{a-b}=f\frac{a+b+c}{x+y+z}}$
Similarly ${(2)-(3)}$ and ${(3)-(1)}$, then we get
${\frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}=f\frac{a+b+c}{x+y+z}}$
I don't know how to proceed further. I couldn't get the other stationary value of ${f}$. I need help in proceeding further to calculate stationary points and/or stationary values, if at all, my method is correct.

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,
$$\begin{align}(f,x,y,z)&=\Biggl(\frac{1}{a^3+b^3+c^3-3abc},\frac{a^2-bc}{a^3+b^3+c^3-3abc}\\&\phantom{aaaaa},\frac{b^2-ca}{a^3+b^3+c^3-3abc},\frac{c^2-ab}{a^3+b^3+c^3-3abc}\Biggr)\,.\end{align}\tag{6}$$
