Verify that $x^2 + cy^2 = 1$ is an implicit solution to $\frac{dy}{dx} = \frac{xy}{x^2 - 1}.$ When I differentiate $x^2 + cy^2 = 1$ implicitly and solve for $\dfrac{dy}{dx}$, I get $\dfrac{dy}{dx} = \dfrac{-x}{cy}.$ I thought I had to multiply by a fraction in order to make it similar to $\dfrac{dy}{dx} = \dfrac{xy}{x^2 - 1}$\$ but nothing worked.
I then checked the answer and someone put up these steps:
\begin{align*}
y^2 &= -(x^2 - 1) * \dfrac{1}{c} \\
y^2 &= c(x^2 - 1) \\
y &= \sqrt{c(x^2 - 1)} \\
y &= c\sqrt{x^2 - 1)}
\end{align*}
I don't get how $\dfrac{1}{c}$ changes to $c$ and it's square root disappears.

Notice that:
$$
x^2 + cy^2 = 1 \iff cy = \frac{1 - x^2}{y}
$$
So:
$$
\frac{dy}{dx} = \frac{-x}{cy} = \frac{-x}{\left(\frac{1 - x^2}{y} \right)} = \frac{-xy}{1 - x^2} = \frac{xy}{x^2 - 1}
$$
