Thanks so much! That is indeed a great solution!
An alternative that I thought of was:
f'(0)^2 = f(0)f''(0) --> Since f(0) = 1, we conclude that f''(0) = f'(0)^2. Now, we can differentiate both sides to get f'(x)f''(x)=f(x)f'''(x). Now, we can differentiate this equation once again! And use the product rule! --> f'(x)f'''(x) + f''(x)^2 = f(x)f''''(x) + f'(x)f'''(x) --> f''(x)^2 = f(x)f''''(x).
Plugging in x=0 simply gives us the answer, that is, f''(0)^4 = f'''(0) = 0. Therefore, f''(0) = \(\pm 3\) and further, since f'(0)^2 =f''(0), we get the end result, \(f'(0) = \pm \sqrt{3}\). Yayyyy!
