an ordinary point
Before you commit to building a power-series solution around some location x0, it pays to ask: is this a 'safe' place to expand? The answer depends on how the equation's coefficients behave right there. An ordinary point is a safe place — a point where the coefficients are well-behaved enough that an ordinary Taylor-series solution is guaranteed to exist and converge.
Make it precise. Write the equation in standard form y'' + P(x) y' + Q(x) y = 0, dividing through so the leading coefficient is 1. A point x0 is an ordinary point if both P(x) and Q(x) are analytic at x0 — meaning each is itself given by a convergent power series there, which for the usual rational coefficients simply means neither blows up (no division by zero) at x0. For example, in y'' + x y' + (x^2 + 1) y = 0 every point is ordinary, because the coefficients are polynomials and never misbehave. At an ordinary point you may confidently set y = sum a_n (x - x0)^n, knowing two independent series solutions exist with the expected free constants a_0 and a_1.
The contrast is a singular point, where P or Q does blow up; there the ordinary method can fail and you need the heavier method of Frobenius. The practical payoff: locating the ordinary points (and the singular ones) up front tells you exactly where you may expand and, through the nearest singularity, how far the resulting series will converge.
In (x^2 - 1) y'' + 2x y' - 6 y = 0, divide by (x^2 - 1) to get P = 2x/(x^2-1), Q = -6/(x^2-1). These blow up only at x = 1 and x = -1, so every other point — in particular x0 = 0 — is an ordinary point and you may expand there safely.
Always divide to standard form first: ordinariness is judged on P and Q, not on the raw leading coefficient.
A point where the leading coefficient vanishes — like x = 1 above — is automatically suspect, but you must check P and Q after dividing, not the original equation, because the trouble lives in the ratios.