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.

@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.
