How to figure out if there is an actual horizontal tangent without a graph There is this practice problem that asks to determine the points at which the graph of $y^4=y^2-x^2$ has a horizontal tangent.
So I did implicit differentiation to find that 
$$\displaystyle\frac{dy}{dx} = \frac{-x}{2y^3-y}$$
To find the horizontal tangent, I set $\frac{dy}{dx}=0$ and solved for $x$:
$$\begin{align}
\displaystyle\frac{dy}{dx} &= 0 \\
\frac{-x}{2y^3-y} &= 0 \\
-x &= 0 \\
x &= 0
\end{align}$$
Then I substituted $x=0$ into the equation of the curve:
$$\begin{align}
y^4&=y^2-(0)^2 \\
0 &= y^4 - y^2\\
0 &= y^2(y+1)(y-1) \\
y&=-1,\,0,\,1
\end{align}$$ 
I concluded that the points $(0,0)$, $(0,-1)$, and $(0,1)$ were the points with a horizontal tangent.
However, when I graphed this using Desmos, it turns out that the point at $(0,0)$ did not look like it has horizontal tangent.
Graph of y^4=y^2-x^2
How would I have been able to figure this out without graphing it?

If
$y^4=y^2-x^2
$,
then
diffing implicitly,
$4y^3y'
=2yy'-2x
$
or
$2y^3y'
=yy'-x
$.
If $y' = 0$,
then
$x = 0$.
Putting this in,
$y^4 = y^2$
so the possible values are
$y = 0, \pm 1$.
At $x=y=0$,
suppose $y' = c$.
For small $x$ and $y$,
$y \approx cx$
so
$c^4x^4 \approx c^2x^2-x^2$
or,
dividing by $x^2$,
$c^4 x^2 \approx c^2-1$.
Since
$c^4x^2$ is small,
$c^2-1$ must also be small
so
$c^2 \approx 1$
so
$c \approx \pm 1$.
This means that,
at $(0, 0)$,
$y' = \pm 1$.
