<s>[INST] if $f'''(x)$ is continuous everywhere and $\lim_{x \to 0}(1+x+ \frac{f(x)}{x})^{1/x}=e^3$ Compute $f''(0)$ 
if $f'''(x)$ is continuous everywhere and $$\lim_{x \to 0}(1+x+ \frac{f(x)}{x})^{1/x}=e^3$$ Compute $f''(0)$

The limit equals to $$\begin{align} \lim_{x \to 0} \frac{\log(1+x+ \frac{f(x)}{x})}{x}-3=0.  \end{align}$$ 
From $$\frac{\log(1+x+ \frac{f(x)}{x})}{x}-3=o(1)$$ as $x \to 0$, I get $$1+x+\frac{f(x)}{x} = e^{3x+o(x)},$$ and  $$f(x)=x(e^{3x+o(x)}-x-1),\frac{f(x)}{x}=e^{3x+o(x)}-x-1$$ as $x \to 0$. So both $f(0)$ and $f'(0)$ are $0$. Approximating $e^{3x+o(x)}=1+3x+o(x)$ I get $$\begin{align} f(x) &= x(1+3x+o(x)-x-1) =2x^2+o(x^2). \end{align}$$
Now I try to use the definition of derivative to calculate the $f''(x)$ 
$$f''(x)=\lim_{x \to 0}\frac{f'(x)-f'(0)}{x}=\lim_{x \to 0} \frac{f'(x)}{x}$$
I'm not sure whether I can get $f'(x)$ by differentiating the approximation $2x^2+o(x^2)$ and how to differentiate $o(x^2)$.
 [/INST] I will use the following formula for $f''(0)$. We have 
$$
f(x)=f(0)+f'(0)x+f''(0)\,\frac{x^2}2+o(x^3),
$$
$$
f(-x)=f(0)-f'(0)x+f''(0)\,\frac{x^2}2-o(x^3).
$$
Adding and solving for $f''(0)$, we get
$$\tag{1}
f''(0)=\frac{f(x)+f(-x)-2f(0)}{x^2}+o(x).
$$
Starting from your 
$$
1+x+\frac{f(x)}{x} = e^{3x+o(x)},
$$
we have
$$\tag{2}
x+x^2+f(x)=xe^{3x+o(x)}.
$$
Then, taking $x=0$, we get  $f(0)=0$;  and, using $(1)$,
\begin{align}
f''(0)&=\lim_{x\to0}\frac{f(x)+f(-x)-2f(0)}{x^2}=\lim_{x\to0}\frac{f(x)+f(-x)}{x^2}\\ \ \\
&=\lim_{x\to0}\frac{xe^{3x+o(x)}-x-x^2-xe^{-3x+o(-x)}+x-x^2}{x^2}\\ \ \\
&=-2+\lim_{x\to0}\frac{e^{3x+o(x)}-e^{-3x+o(-x)}}{x}\\ \ \\
&=-2+\lim_{x\to0}\frac{3x+3x+o(x^2)}{x}\\ \ \\ &=-2+6=4.
\end{align}
In the last limit I'm cancelling $o(x)$ with $o(-x)$; this is ok since both terms are coming from the same expression $(2)$.
</s>