I hope it's not too late
for 1) I'm assuming you mean lim (log(n^2))/n. If this is the case, you can take the log(n^2) and change it to 2*log(n). After that, we see that n > log(n) as n->infinity. Therefore, limit = 0.
2) as n->infinity, 3/n -> 0. Then, 1^n = 1. Thus, limit as n->infinity = 1.
3) I'd use the Alternating Series test. This gives us two criteria to find: is An+1 < An, and is lim An = 0? Both of these are satisfied, so it is convergent.
4) We rearrange this a bit:
-3x^2*y*ln(y) + (2*y - x^3 - x^3*ln y)*(dy/dx) = 0.
An exact equation is in the form P(x) + Q(x)(dy/dx) = 0.
P(x) = -3x^2*y*ln(y) and Q(x) = 2*y - x^3 - x^3*ln(y)
To find out if the equation is exact, the partial derivative of P(x) wrt y must equal the partial of Q(x) wrt x.
Py = -3*x^2*ln(y) - 3*x^2
Qx = -3*x^2-3*x^2*ln(y)
Py = Qx, therefore the equation is exact.