<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] Here is a solution that necessitates to have had a previous introduction

*

*to conic curves (maybe it is the case) and to


*barycentric coordinates (less likely, unless you have prepared some Mathematics Olympiads) ; here or here are a nice introduction to barycentric coordinates (b.c. in short). One of the most elementary (and important) property is that b.c. are defined up to a multiplicative factor.
Indeed, for a fixed triangle of reference, your issue amounts to assert the bijectivity (one-to-one onto) of the following transformation,
$$M\underbrace{(x, \ y, \ z)}_{\text{b.c.}} \ \ \xrightarrow{(H)} \ \ M'\underbrace{(x^2-yz,\ \ y^2-zx, \ \ z^2-xy)}_{\text{b.c.}} \tag{1}$$
known as Hirst transform (see here).
This transformation can be defined geometrically: $M'$ is the intersection point of line $GM$ (where $G$ is the center of mass of the triangle) and polar line of $M$ with respect to the Steiner interior ellipse. See figure below and proof in the Appendix at the bottom.

Fig. 1: Steiner ellipse (S), tangent to the triangle's sides in their midpoints. $G$ is the center of mass. Being given point $M$, its Hirst transform is $M'=(P) \cap GM$, where $(P)$ is the polar line of $M$ with respect to $(S)$ (in green). Please note that in this case of figure where $M$ is outside the ellipse, (P) has been obtained by connecting the tangency points of the tangents to (S) issued from $M$.
Knowing $M'$, it is easy to retrieve its unique preimage $M$, establishing the bijectivity of Hirst transform $(H)$, even more its involutivity $(H)^{-1}=H$. This involutivity can be checked, of course, in an algebraic way.
Remarks :

*

*Strictly speaking, we have to exclude the case where $M$ has identical barycentric coordinates $(x,y,z)$ with $x=y=z$, which means that $M$ is situated at the center of mass $G$ ; indeed, in this case, line $GM$ is undefined. It is a confirmation of the exceptional "counterexample" $x=y=z=1$ found by @Macavity.


*See the very interesting analysis in the other answer by @orangeskid.


*Hirst transform is mentioned but in a handful of texts such as this one in connection with the study of certain cubic curves.


*The "nonclassical" life of Mr Hirst is described here.
Appendix: Proof of the geometrical interpretation:
The internal Steiner ellipse has equation:
$$x^2+y^2+z^2-2(xy+yz+zx)=0$$
which is equivalent to:
$$2(xy+yz+zx)=\frac12\tag{2}$$
(see here p. 119). Therefore, the polar of point $(x,y,z)$, is given by the bilinear expression associated with quadratic form in the LHS of (2) has equation:
$$(x'y+y'x)+(y'z+z'y)+(z'x+x'z)=\frac12 \iff $$
$$x'(y+z)+y'(z+x)+z'(x+y)=\frac12\tag{3}$$
Besides, the equation of line $GM$ is:
$$\begin{vmatrix}1&x&x'\\1&y&y'\\1&z&z'\end{vmatrix}=0 \iff x'(z-y)+y'(x-z)+z'(y-x)=0\tag{4}$$
The intersection of straight lines given by (3) and (4) is obtained by solving the system (3)+(4), giving, up to a multiplicative constant (as always with b.c.) :
$$\begin{pmatrix}x'\\y'\\z'\end{pmatrix}=\begin{pmatrix}-x^2+yz\\-y^2+zx\\-z^2+xy\end{pmatrix}$$
as awaited.
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