Determine the singular points of the following differential equation and classify each singular point as regular or irregular.

(x^{2} -1)y"+3(x+1)y'+y=0

Let be
p(x)=(x^{2} -1
q(x)=3(x+1)
r(x)=1
the three coefficients of the equation
a is a singular point of the equation if   lim p(x) =0
                                                              x>a 
so let's find a
 lim p(x) = lim x²-1=a²-1=0
  x>a    x>a 
a²-1=0 implies  a=+ or -1

so the sigular points are a= -1 or a=1

case 1
for a= -1
lim (x-(-1)) q(x)/p(x)=lim (x+1) 3(x+1)/x²-1=lim3(x+1)/x-1= 0/-2=0
x> -1                  x> -1                    x> -1
lim (x-(-1))²  r(x)/p(x)= lim(x+1)²/x²-1= 0/-2=0
x> -1                    x> -1

lim (x-(-1)) q(x)/p(x) and lim (x-(-1))²  r(x)/p(x) are finite so  -1 is regular 
 x> -1                        x> -1
singular point

case 2
a=1
lim (x-1)) q(x)/p(x)=lim (x-1) 3(x+1)/x²-1=lim3(x+1)/x+1= 3
x> 1                  x> 1                    x> 1
lim (x-1))²  r(x)/p(x)= lim(x-1)²/x²-1= =0
x> 1                    x> 1

1 is also a regular singular point