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.
 A: 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.
A: An idea would be to show that if
$$[p\colon q\colon r] = [x^2 - y z \ \colon y^2 - x z\  \colon z^2 - x y]$$
then
$$[x\ \colon y \colon z] = [p^2 - q r\ \colon q^2 - p r\  \colon r^2 - p q]$$
In other words, the map from $\mathbb{P}^2$ to itself given by
$$[x\ \colon y \colon z]\mapsto [x^2 - y z \ \colon y^2 - x z\  \colon z^2 - x y]$$
is an involution.
This can be checked by a direct calculation. Alternatively, we can see that for the matrix
$$A = \left (\begin{matrix} x & y & z \\ y & z & x \\ z & x & y\end{matrix} \right)$$
the adjugate matrix is of similar form
$$\operatorname{adj} A = \left (\begin{matrix} x^2 - y z  & y^2 - x z  & z^2 - x y \\ y^2 - x z  & z^2 - x y & x^2 - y z \\ z^2 - x y  & x^2 - y z  & y^2 - x z \end{matrix} \right)$$
and use the formula
$$\operatorname{adj}\operatorname{adj}A = \det A \cdot A $$
Notice that
$$\det A = \frac{1}{2}(x+y+z) ((x-y)^2 + (y-z)^2 + (z-x)^2)$$
so according to the conditions in OP, it is $\ne 0$.
$\bf{Added:}$
Some calculations related to Hirst transform. Given a quadric $Q(p)=0$ and a point $p_0$ not on the quadric, to consider the "inversion" wr to $Q$ and center $p_0$. It is like the inversion wr to a sphere, but with a projective description. I am following @Jean Marie:
For a point $p$ in projective space, consider the intersection of the line $p_0p$ with the hyperplane polar to $p$. Denote by $\langle \cdot,\cdot\rangle$ the bilinear form giving $Q$. We have $\phi(p) = \alpha p_0 + \beta p$, such that
$$\alpha p_0 + \beta p \perp p$$
So we can take
$$\phi(p) = \langle p,p \rangle  p_0 - \langle p, p_0 \rangle p $$
Check that
$$\langle \phi(p), p \rangle = 0 \\
\langle \phi(p), p_0\rangle = G(p, p_0)\\
 \langle \phi(p), \phi(p)\rangle = G(p, p_0) \cdot \langle p, p\rangle $$
where $G(p, p_0)$ is the Gram determinant of $p$, $p_0$, which gives
$$\phi\circ \phi(p) = \langle p, p_0 \rangle\cdot G(p, p_0) \cdot p$$
so projectively $\phi$ is an involution.
A: @Jean Marie and @orangeskid gave very nice proofs.
Their idea, in algebraic manner, leads to the following proof:
We have the following identity
$$(p^2 - qr)^2 - (q^2 - rp)(r^2 - pq)
= p(p + q + r)(p^2 + q^2 + r^2 - pq - qr - rp).$$
Using this identity, we have
$$\frac{(x^2 - yz)^2 - (y^2 - zx)(z^2 - xy)}{(a^2 - bc)^2 - (b^2 - ca)(c^2 - ab)}
= k\,\frac{x}{a}, \tag{1}$$
and
$$\frac{(y^2 - zx)^2 - (z^2 - xy)(x^2 - yz)}{(b^2 - ca)^2 - (c^2 - ab)(a^2 - bc)}
= k\, \frac{y}{b}, \tag{2}$$
and
$$\frac{(z^2 - xy)^2 - (x^2 - yz)(y^2 - zx)}{(c^2 - ab)^2 - (a^2 - bc)(b^2 - ca)}
= k \, \frac{z}{c}, \tag{3}$$
where
$$k = \frac{(x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)}{(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)}.$$
Note: One can prove $k \ne 0$.
See the remarks at the end.
We are done.

Remarks:
If $k = 0$, we have $x + y + z = 0$.
Using $z = -x - y$, we have
$$\frac{x^2 + xy + y^2}{a^2-bc}=\frac{x^2 + xy + y^2}{b^2-ca}=\frac{x^2 + xy + y^2}{c^2-ab} \ne 0$$
which results in
$a^2 - bc = b^2 - ca = c^2 - ab$
and thus $(a-b)(a + b + c) = 0$ and $(b - c)(a + b + c) = 0$.
Thus, $a = b = c$ which contradicts the given conditions.
A: 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.
