Confusing solution to the limit of an implicit function? $$\frac{8}{3}=\frac{\log{x}}{\log{y}}-\frac{\log{y}}{\log{x}}$$
When I graphed this implicit function on desmos (https://www.desmos.com/) it appeared as if there were two solutions as $x\to{0}$ from the positive direction: $y\to\infty$ and $y\to{0}$. However, neither of these solutions make much sense to me. 
Are these correct? If they are please could you show me an algebraic way to solve this? If not can you please explain why this is incorrect? 
 A: Let $z=\frac{\log x}{\log y }$.  Then, your equation becomes
$$
\frac{8}{3}=z-\frac{1}{z}
$$
Then, we multiply by $z$ to get
$$
z^2-\frac{8}{3}z-1=0.
$$
We can solve this to get
$$
z=\frac{\frac{8}{3}\pm\sqrt{\frac{64}{9}+4}}{2}=\frac{\frac{8}{3}\pm\frac{10}{3}}{2}.
$$
In other words, $z=3$ or $z=\frac{-1}{3}$.  Therefore, the solutions to this implicit formula obey either $\frac{\log x}{\log y}=3$ or $\frac{\log x}{\log y}=-\frac{1}{3}$.
Now, you can study each case separately.  In particular, when $z=3$, you know that
$$
\frac{\log x}{\log y}=3 
$$
or that 
$$
\frac{1}{3}\log x=\log y.
$$
Taking the exponential of both sides,
$$
y=x^{1/3}.
$$
As $x$ approaches $0$, this cube root also approaches $0$.
On the other hand, when $z=-\frac{1}{3}$, we have that
$$
-3\log x=\log y.
$$
Taking the exponential of both sides,
$$
y=x^{-3}.
$$
As $x$ approaches $0$, the cube of the reciprocal approaches $\infty$.
This confirms your original statement.
At this point, we can argue if there is a limit as $x$ approaches $0$.  One could not say $\lim_{x\rightarrow 0}y(x)$ exists because $y$ is not a function of $x$.  However, if you look locally, then there are two branches and on each branch the limit as $x$ approaches $0$ is well defined.
