How to find the particular solution of the differential equation. Original equation $$x^8y'+7y=e^\frac{1}{x^7}\quad y(1)=e$$ This is the original equation with the initial condition. How would i separate to take the integral? This is what I have done so far but I am not sure if it right and if it is I do not know how to continue.
$$y'+\frac{7y}{x^8}=\frac{e^\frac{1}{x^7}}{x^8}$$
Your steps are correct so far:
$$ \frac{dy}{dx} + 7x^{-8}y = e^{x^{-7}}x^{-8}$$

You can solve this using the method of integrating factors:

Your integrating factor (IF) is:

$$ \text{IF} = e^{\int 7x^{-8}}dx = e^{-x^{-7}}$$

You then multiply through by IF:

$$ e^{-x^{-7}}\frac{dy}{dx} + 7x^{-8}ye^{-x^{-7}} = x^{-8} $$

$$  \frac{d}{dx}\left(ye^{-x^{-7}}\right) = x^{-8}$$

Integrating both sides wrt $x$:

$$ ye^{-x^{-7}} = \int x^{-8}dx = -\frac{1}{7}x^{-7} + c$$

where $c$ is a constant.

$$ y = -\frac{1}{7}x^{-7}e^{x^{-7}} + ce^{x^{-7}}$$

You can then use $y(1) = e$ to find an expression for $c$ and hence a solution to the differential equation.
$$y'+\frac{7y}{x^8}=\frac{e^\frac{1}{x^7}}{x^8}$$

If $y' + P(x)y = q(x)$ then $y=e^{-\int p(x) dx}(\int e^{\int p(x) dx}q(x) dx+c) $ is answer. thus

$$y=e^{\int \frac{-7}{x^8}}(\int e^{\int \frac{7}{x^8}}\frac{e^\frac{1}{x^7}}{x^8}+c)=e^{{ \frac{1}{x^7}}}(\int e^{{ \frac{-1}{x^7}}}\frac{e^\frac{1}{x^7}}{x^8}+c)=e^{{ \frac{1}{x^7}}}(\frac{-1}{7x^7}+c)$$ since $y(1)=e$ so $e=e^{{ \frac{1}{1^7}}}(\frac{-1}{7}+c)\Rightarrow c=\frac{8}{7}$