What is positive-0 squared minus positive-0? I've got a basic limit problem that I think I'm solving the right way, but I've run into something that looks confusing enough to make me wonder if I'm doing it right.

$$
\lim_{y\to0} \frac{1}{y^2-y} + \frac{1}{y}
$$

I tried approaching this by looking at the Left Hand Limit and Right Hand Limit. Though when I plug in $0^+$, I get

$$
\lim_{y\to0} \frac{1}{0^+-0^+} + \frac{1}{0^+}
$$

What is $0^+-0^+$? Is it still positive? or is it just 0, or in this case, would it be negative? I'm leaning toward negative because for all $0\lt y\lt 1$, $y^2$ will be less than $y$, so then wouldn't I have an infinitely small negative number as a result?

And then, if that turns out to be true, it would appear that this problem is unsolvable by that method then, seeing as I'd have $+\infty + -\infty$, which is undefined.
Clearly $y^2-y\to 0$ as $y\to 0$, as does $y$, so both fractions are blowing up, and their difference is an indeterminate form. Combine them over a common denominator; if you do the algebra correctly, you’ll be able to simplify the resulting fraction to one whose limit as $y\to 0$ is easy to compute.
It's $-1$, silly!  Obviously, you're very confused based on your argument.  I suggest coming back at the problem with fresh eyes after you've taken a break from it.