I will run through two methods: the 'unofficial' method, and the (more) 'official' method.
Intuitive approach around the limit:
Using the compound angle and double angle formulas for sin and cos:
sin(3y) = sin (2y + y)
=sin(2y)cosy + cos(2y)siny)
=2sinycos^2y+(2cos^2y-1)siny
=4sinycos^2y - siny
(Rearranging to get it all in terms of sin(y))
=4siny(1-sin^2y) - siny
=3siny - 4sin^3y
Now we want to find the square of that (sin(3y)):
sin^2(3y) = (3siny - 4sin^3y)^2
=(siny(3-4sin^2y))^2
=sin^2y(3- (4 - 4cos^2y))^2
= sin^2y(4cos^2y - 1)^2
Now to find:
lim_(y rarr 0) (sin^2y(4cos^2y - 1)^2 - 3y^2)/(y^2 + y^3)
This is where we start to get a bit 'intuitive'...
As y rarr 0, siny rarr 0
rArr siny rarr y
Hence let y = siny.
Meanwhile cosy rarr 1
Hence let cosy = 1.
This simplifies the limit down to:
lim_(y rarr 0) (y^2(3)^2 - 3y^2)/(y^2 + y^3)
=lim_(y rarr 0)(6y^2)/(y^2 + y^3)
=6 lim_(y rarr 0) y^2/(y^2(1+y))
=6 lim_(y rarr 0) 1/(1+y)
Finally letting y = 0 we get:
6 lim_(y rarr 0) 1
= 6
Official approach to the limit:
This uses L'Hopital's Rule, i.e. if lim_(y rarr 0) f(x)/g(x) yields a mathematically impossible result, then the limit will be the same if you repeatedly differentiate f and g to the point where the result is no longer undefined.
lim_(y rarr 0) (sin^2(3y) - 3y^2)/(y^2 + y^3)
lim_(y rarr 0) (d/dx(sin^2(3y) - 3y^2))/(d/dx(y^2 + y^3))
By the chain rule, this is equal to:
lim_(y rarr 0) (6sin(3y)cos(3y) - 6y)/(2y + 3y^2)
Still gives out the zero denominator, so we do it all again!
By the product and chain rules:
d/dx(6sin(3y)cos(3y)
=6(3cos(3y)cos(3y) + (-sin(3y)sin(3y))
=6(3cos^3y - 3sin^3y)
Hence:
lim_(y rarr 0) (6sin(3y)cos(3y) - 6y)/(2y + 3y^2)
=lim_(y rarr 0) (18cos^2(3y) - 18sin^2(3y) - 6)/(2 +6y)
Rearranging the trigonometric parts of the numerator using the Pythagorean Identity, we get:
lim_(y rarr 0) (36cos^2(3y) - 18 - 6)/(2 +6y)
Finally letting y=0, we get:
(36-18-6)/2
=6
Goodness, that was long - I hope it helps!