Power Series Solution for $e^xy''+xy=0$ $$e^xy''+xy=0$$
How do I find the power series solution to this equation, or rather, how should I go about dealing with the $e^x$? Thanks!

When trying to find a series to represent something, it's important to decide what kind of a series you want.  Even if you're dealing with power series, a small change in notation between $\displaystyle \sum a_n x^n$ and $\displaystyle\sum \frac{a_nx^n}{n!}$ can lead to substantial changes.  In particular, we have that, if $f(x)=\sum a_n x^n$, then 
$$e^x f(x) = \left(\sum \frac{x^n}{n!}\right) \left(\sum a_n x^n\right) = \sum \left(\sum_{i+j=n}\frac{a_i}{j!} \right) x^n, $$
which, while it will lead to a perfectly good recurrence yielding power series solution for a problem like this, is somewhat awkward, unwieldy, and likely not to lead to a recurrence that you can recognize or explicitly solve.  However, if $f(x)=\sum \frac{a_n}{n!}$, then 
$$e^x f(x) = \left(\sum \frac{x^n}{n!}\right) \left(\sum \frac{a_n x^n}{n!}\right) = \sum \left(\sum_{i+j=n}\frac{a_i n!}{i!j!} \right) \frac{x^n}{n!}=\sum \left(\sum_{k\leq n}a_k \binom{n}{k} \right) \frac{x^n}{n!}, $$
which is a nicer looking expression.  Additionally, the power series expansion of  $f'(x)$ has a nice form due to the cancellation between the $n$ in $n!$ and the $n$ in $D(x^n)=nx^{n-1}$
With this in mind, I suggest a slight change on the hint of  @M.Strochyk
Hint: Expand $\displaystyle y(x)=\sum \frac{a_nx^n}{n!}$ and $\displaystyle e^x=\sum \frac{x^n}{n!}$ and substitute them into the equation.
