Use the quotient rule for differentiation, (dy)/(dx) = (v du)/(dx) - u (dv)/(dx)v², where u = e⁸x and v = 1+e⁴x.

First, find the derivatives of $u$ and $v$:

$$\frac{du}{dx} = \frac{d}{dx}(e^{8x}) = 8e^{8x}$$ $$\frac{dv}{dx} = \frac{d}{dx}(1+e^{4x}) = 4e^{4x}$$

$$\frac{dy}{dx} = \frac{8e^{8x} + 8e^{12x} - 4e^{12x}}{(1+e^{4x})^2}$$ $$\frac{dy}{dx} = \frac{8e^{8x} + 4e^{12x}}{(1+e^{4x})^2}$$

Substitute $x=0$ into the derivative:

$$\frac{dy}{dx}\Big|_{x=0} = \frac{8e^{8(0)} + 4e^{12(0)}}{(1+e^{4(0)})^2}$$ $$\frac{dy}{dx}\Big|_{x=0} = \frac{8e^0 + 4e^0}{(1+e^0)^2}$$

Since $e^0 = 1$:

$$\frac{dy}{dx}\Big|_{x=0} = \frac{8(1) + 4(1)}{(1+1)^2} = \frac{8+4}{2^2} = \frac{12}{4} = 3$$

The condition $y=1$ at $x=0$ is not consistent with the given function $y(0) = \frac{e^0}{1+e^0} = \frac{1}{2}$. We proceed by evaluating the derivative at $x=0$.
The final answer is $\boxed{3}$.

2. Solve $\frac{dy}{dx}$ if $y = \ln\left(\frac{x+1}{x-1}\right)$.

Recall that $\frac{d}{dx}(\ln u) = \frac{1}{u} \frac{du}{dx}$.

$$\frac{dy}{dx} = \frac{d}{dx}(\ln(x+1)) - \frac{d}{dx}(\ln(x-1))$$ $$\frac{dy}{dx} = \frac{1}{x+1} \cdot \frac{d}{dx}(x+1) - \frac{1}{x-1} \cdot \frac{d}{dx}(x-1)$$ $$\frac{dy}{dx} = \frac{1}{x+1}(1) - \frac{1}{x-1}(1)$$ $$\frac{dy}{dx} = \frac{1}{x+1} - \frac{1}{x-1}$$

$$\frac{dy}{dx} = \frac{(x-1) - (x+1)}{(x+1)(x-1)}$$ $$\frac{dy}{dx} = \frac{x-1-x-1}{x^2-1}$$ $$\frac{dy}{dx} = \frac{-2}{x^2-1}$$

The final answer is $\boxed{\frac{-2}{x^2-1}}$.

3. Find intercepts and asymptotes for $h(x) = \frac{2x-4}{x^3-1}$.

a) x-intercepts:
Set $h(x) = 0$. This occurs when the numerator is zero.

$$2x-4 = 0$$ $$2x = 4$$ $$x = 2$$

The x-intercept is $\boxed{(2, 0)}$.

b) y-intercepts:
Set $x = 0$.

$$h(0) = \frac{2(0)-4}{0^3-1} = \frac{-4}{-1} = 4$$

The y-intercept is $\boxed{(0, 4)}$.

c) Vertical asymptotes:
Set the denominator to zero.

$$x^3-1 = 0$$

Factor the difference of cubes: $(x-1)(x^2+x+1) = 0$.
The real root is $x-1=0 \implies x=1$.
The quadratic factor $x^2+x+1$ has a discriminant $\Delta = 1^2 - 4(1)(1) = -3 < 0$, so it has no other real roots.
Check that the numerator is not zero at $x=1$: $2(1)-4 = -2 \ne 0$.
The vertical asymptote is $\boxed{x=1}$.

d) Horizontal asymptotes:
Compare the degrees of the numerator and denominator.
Degree of numerator ($N$) = 1.
Degree of denominator ($D$) = 3.
Since $N < D$, the horizontal asymptote is $y=0$.
The horizontal asymptote is $\boxed{y=0}$.

e) Slant asymptotes:
A slant asymptote exists if the degree of the numerator is exactly one more than the degree of the denominator ($N = D+1$). In this case, $N=1$ and $D=3$, so $N \ne D+1$.
There is $\boxed{\text{no slant asymptote}}$.

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