<s>[INST] If $\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab}$, prove that $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$. 
Problem: Let $a, b, c, x, y, z$ be real numbers
such that $abc \ne 0$, $ ~ a + b + c \ne 0$, $~ (a^2 - bc)(b^2 - ca)(c^2 - ab)\ne 0$, $xyz\ne0$, $x+y+z\ne0$, $(x-y)^2+(y-z)^2+(z-x)^2\ne0$
and $$\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab} \ne 0.$$
Prove that $$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}.$$

The converse can be proved easily.
My Attempt:
$
\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab}\\
\Rightarrow\frac{x^2-yz-y^2+zx}{a^2-bc-b^2+ca}=\frac{y^2-zx-z^2+xy}{b^2-ca-c^2+ab}=\frac{z^2-xy-x^2+yz}{c^2-ab-a^2+bc}\ [By\ Addendo]\\\Rightarrow\frac{x-y}{a-b}\cdot\frac{x+y+z}{a+b+c}=\frac{y-z}{b-c}\cdot\frac{x+y+z}{a+b+c}=\frac{z-x}{c-a}\cdot\frac{x+y+z}{a+b+c}\\\Rightarrow\frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}\ \left[Considering\ (x+y+z)\ and\ (a+b+c)\ \neq0\right]
$
Known Exceptional Counter-examples:-
Macavity: $a=1, b=2, c=3, x=y=z$
River Li: $a = 1, b=2, c = -3, x = 7, y = 7, z = -14$
The conditions in the question have been edited to exclude exceptional counter-examples.
 [/INST] Proof:
First of all, clearly we have
$$ (a - b)^2 + (b - c)^2 + (c - a)^2 \ne 0. \tag{1}$$
Now, we split into two cases:
Case 1  $~ x = y$:
We have
$$\frac{x^2-xz}{a^2-bc}=\frac{x^2-zx}{b^2-ca}=\frac{z^2-x^2}{c^2-ab} \ne 0$$
which results in
\begin{align*}
 a^2 - bc &= b^2 - ca, \tag{2}\\
 x(c^2 - ab) &= - (z + x)(b^2 - ca). \tag{3}
\end{align*}
From (2), we have $(a - b)(a + b + c) = 0$
which results in $a = b$.
With this and (3), we have
$x(c^2 - a^2) = - (z + x)(a^2 - ca)$ or $(a - c)(az - cx) = 0$ which results in $az - cx = 0$ or $\frac{x}{a} = \frac{z}{c}$,
where we have used $a = b$ and (1) to get $a \ne c$.
Thus, we have
$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$.
Case 2 $~ x \ne y$:
Let
\begin{align*}
 f &= (a^2 - bc)(y^2 - zx) - (b^2 - ca)(x^2 - yz) \\
 &= Az + B,\\
 g &= (a^2 - bc)(z^2 - xy) - (c^2 - ab)(x^2 - yz) \\
 &= Cz^2 + Dz + E
\end{align*}
where
\begin{align*}
 A &= -{a}^{2}x-acy+{b}^{2}y+bcx, \\
 B &= {a}^{2}{y}^{2}+ac{x}^{2}-{b}^{2}{x}^{2}-bc{y}^{2}, \\
 C &= a^2 - bc,\\
 D &= -aby+{c}^{2}y, \\
 E &= -{a}^{2}xy+ab{x}^{2}+bcxy-{c}^{2}{x}^{2}.
\end{align*}
We eliminate $z$ from the system of $f = g = 0$ as follows.
Let
$$h = Cz \cdot f - A\cdot g = (BC - AD)z - AE.$$
Then let
\begin{align*}
 F &=  (BC - AD)\cdot f - A \cdot h\\
  &= 
 A^2 E - ABD + B^2C\\
 &= - \frac18 (x - y) [3(x + y)^2 + (x - y)^2]  ( ay - bx )  ( a + b + c )  \\
 &\qquad \times ( a^2 - bc ) [(a - b)^2 + (b - c)^2 + (c - a)^2].
\end{align*}
From $F = 0$, using (1) and the given conditions, we have $ay - bx = 0$
or $\frac{x}{a} = \frac{y}{b}$.
Since $x \ne y$, we have $a \ne b$.
Using $x = \frac{a}{b}y$,
we have
$$f = - \frac {y (a - b)  [ 3(a + b)^2 + (a - b)^2] ( bz - cy) }{4 b^2} = 0
$$
which results in $bz = cy$ or $\frac{y}{b} = \frac{z}{c}$.
Thus, we have
$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$.
We are done.
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