Let g(x) = 3x^5+1 and h(x) = 2x^6+3x-1
f(x)=(3x^5+1)/(2x^6+3x-1) = g(x)/(h(x))
g(x) is of degree 5 and h(x) is of degree 6, so:
f(x) = g(x)/(h(x))->0 as x->+-oo
So f(x) has horizontal asymptote y = 0
f(x) will have vertical asymptotes wherever h(x) = 0 and g(x) != 0
h(x) = 2x^6+3x-1 has 2 changes of sign, so 1 or 2 positive zeros.
h(-x) = 2x^6-3x-1 has 1 change of sign, so h(x) has one negative zero.
Since the coefficients of h(x) are Real, any non-Real zeros will occur in Complex conjugate pairs, so there are an even number of non-Real zeros.
So we can deduce that h(x) has exactly one positive, one negative and four non-Real zeros.
h(x) = 0 has no solution expressible in terms of nth roots, but we can find good approximations using Newton's method.
h'(x) = 12x^5+3
Starting with an initial approximation a_0, iterate using the formula:
a_(i+1) = a_i - (h(a_i))/(h'(a_i)) = a_i - (2a_i^6+3a_i-1)/(12a_i^5+3)
Putting this into a spreadsheet and using initial approximations a_0=1 and a_0=-1, I got the following approximations after a few iterations:
x_1 ~~ 0.3324335502431692846
x_2 ~~ -1.1414892917449508403
Note also that the only Real zero of g(x) is when:
x=root(5)(-1/3)~~-0.80274
So neither of the zeros of h(x) are zeros of g(x) and both of the zeros of h(x) are vertical asymptotes of f(x).
graph{(3x^5+1)/(2x^6+3x-1) [-10, 10, -5, 5]}
