Find a function $f(x)$ so that $f(x)=f'(x)$ I know the answer is $e^x$, but supposing I have no idea, what would be the steps required to get to that answer? Note that starting to test different functions is not a valid answer here as there are infinitely many.

I started by doing 
$$
f(x)=\frac{d}{dx}f(x)
$$
so
$$
\int f(x)dx=f(x)
$$
and 
$$\int f(x)dx=\frac{d}{dx}f(x)$$
$$f(x)=\frac{d^2}{dx^2}f(x)$$
That leads us to
$$\frac{d^n}{dx^n}f(x)=\frac{d^m}{dx^m}f(x)$$
But that gives no new information...

Consider the function $\phi(x) = e^{-x} f(x)$, then show that $\phi' = 0$
and from that conclude that $f(x) = \phi(0) e^x$.
Git Gud asks a good question in the comments. Why $e^x$? Here is one way
of guessing: Suppose that $f$ has a power series expansion around zero and
see what that might be. Note that $f^{(k+1)} = f^{(k)}$, so we have $f'(0) = f(0), f''(0) = f'(0)$, etc. Then we have
$f(x) = f(0) \sum_k {k^k \over k!} = f(0) e^x$. Now check it satisfies the
ODE.
Note that we can write the above as $(D-I) f = 0$, where $D$ is the differential operator. In general (for linear, time invariant, etc. systems), we can write the operator in the form $p(D) f = 0$, where $p$ is a
polynomial. Then we look for a solution of the form $f(x) =  e^ {\alpha x}$
and notice $(p(D) f)(x) = p(\alpha) f (x)$ and we see that $f$ solves the
system iff $p(\alpha) = 0$. In the above case, we have
$p(x) = x-1$, so we try $\alpha = 1$.
