Can you help me solve this algebra problem?

Denote $k=xyz$ and $b$ the common value of $x-\frac{ayz}{x^2}=y-\frac{azx}{y^2}=z-\frac{axy}{z^2}$. We can see that the equation $$ t-\frac{ak}{t^3}=b\tag{*} $$ is satisfied by $t=x,y,z$. Hence $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a polynomial equation of degree 4 $$ t^4-b t^3-ak=0. $$ By Vieta's formula, the other root $w$ satisfies $$ x+y+z+w=b,\quad xyzw=-ak. $$ Since $k=xyz\ne 0$ it follows that $$ w=-a=b-x-y-z, $$ hence we get $$ x-\frac{ayz}{x^2}= y-\frac{azx}{y^2}=z-\frac{axy}{z^2}=b=x+y+z-a. $$

$$x-\frac{ayz}{x^2}=y-\frac{axz}{y^2}$$ gives $$x-y+az\left(\frac{x}{y^2}-\frac{y}{x^2}\right)=0$$ or $$1+\frac{az(x^2+xy+y^2)}{x^2y^2}=0.$$ Similarly, $$\frac{ax(y^2+yz+z^2)}{y^2z^2}=-1$$ and $$\frac{ay(x^2+xz+z^2)}{x^2z^2}=-1.$$ Thus, $$x^3(y^2+yz+z^2)=y^3(x^2+xz+z^2)$$ or $$(x-y)(x^2y^2+xyz(x+y)+z^2(x^2+xy+y^2))=0$$ or $$\sum_{cyc}(x^2y^2+x^2yz)=0$$ or $$\sum_{cyc}z^2(x+y)^2=0,$$ which gives $$x+y=x+z=y+z=0$$ or $$x=y=z=0,$$ which is impossible.

Id est, the given is wrong, which says that $$x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2}\Rightarrow x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2} = x + y + z - a$$ is true.