Note that you are introducing a lot of "false positives". The function implicitly defined by $\frac{2x}{y} - \frac{3y}{x} = 8$ is not the same as the function implicitly defined by $2x^2 - 3y^2 = 8xy.$ Why not? Well, the second function includes $y(0)=0$ (that is, the point $(0,0)$), but the first one is not defined for $x=0$ or $y=0$! Now, it is true that any point on the second curve in which $x\neq 0$ and $y\neq 0$ will be on your original curve, and that the only point in the second curve not on the orginal one is $(0,0)$, so you can make some progress.
These are things you must be careful with when you perform that kind of easy manipulations with implicit functions.
Now, what you seem to be forgetting is that the values of $x$ and of $y$ are connected by the equation $\frac{2x}{y}-\frac{3y}{x} = 8$. So it is entirely possible that the formula $\frac{y}{x}$ will give the exact same values as $\frac{4x-8y}{8x+6y} = \frac{2(x-y)}{4x+3y}$ for any $(x,y)$ that satisfies $\frac{2x}{y}-\frac{3y}{x}=8$. Let's see if that's the case.
We must assume that $x\neq 0$ and that $y\neq 0$; in addition, your formula has the potential unpleasant side effect of also prohibiting points on the line $4x+3y=0$. Are there any points on that line that are on the original curve? If $y = -\frac{4}{3}x$, then plugging into the left hand side of your original equation we have $-\frac{6x}{4x} + \frac{4x}{x} = -\frac{6}{4}+4 \neq 8$ so, phew!, none of the points your formula excludes are on the original graph.
Now, some manipulation is in order. Since $x\neq 0$, $y\neq 0$, and $4x+3y\neq 0$, we have: \begin{align*} \frac{y}{x} = \frac{2x-4y}{4x+3y} &\Longleftrightarrow y(4x+3y) = (2x-4y)x\\ &\Longleftrightarrow 4xy + 3y^2 = 2x^2 - 4xy\\ &\Longleftrightarrow 8xy = 2x^2 - 3y^2\\ &\Longleftrightarrow 8 = \frac{2x}{y} - \frac{3y}{x}. \end{align*} So not only are the two expressions equal on the curve you are working on, they are equal only on the curve you are working on. So the two answers are equivalent, though yours is by far more complicated.