Consider the Taylor's expansion around $c=0$ and find the first 4 terms for the function $\exp(\sin x)$.
I have done this but I'm not sure if is correct.
$$e^{\sin x}=\left(1+x+\frac{x^2}{2!}+\dots+\frac{x^{n-1}}{(n-1)!}\right)^{x-\frac{x^3}{3!}+\frac{x^5}{5!}+\dots+(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!}}$$
On
We may avoid the use of the definition, by taking the composition of the expansions of $e^x$ and $\sin(x)$ at $x=0$: $$\begin{align} e^{\sin(x)}&=e^{x-\frac{x^3}{3!}+o(x^4)}\\ &=1+\left(x-\frac{x^3}{3!}+o(x^4)\right) +\frac{1}{2!}\left(x-\frac{x^3}{3!}+o(x^4)\right)^2 \\&\qquad+\frac{1}{3!}\left(x+o(x^2)\right)^3+\frac{1}{4!}\left(x+o(x^2)\right)^4+o(x^4)\\ &=1+x-\frac{x^3}{6}+\frac{x^2}{2}-\frac{x^4}{6}+\frac{x^3}{6} +\frac{x^4}{24}+o(x^4)\\ &=1+x+\frac{x^2}{2}-\frac{x^4}{8}+o(x^4). \end{align}$$
On
You know that$$e^y=1+y+\frac12y^2+\frac16y^3+\frac1{24}y^4+\cdots\tag1$$and that$$\sin(x)=x-\frac16x^3+\cdots\tag2$$So, in the RHS of$(1)$ replace $y$ with the RHS of $(2)$ and then forget the monomials whose degree is greater than $4$.
On
You want terms up to $x^3$. The exponential is $e^t=1+t+\frac{t^2}2+\cdots$ and the sine $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$. Hence you only need to expand from two terms of the sine, as the others already exceed the degree $3$.
Now, only expanding the necessary terms
$$\begin{align}1&\to1\\ t&\to x-\frac{x^3}6\\ \frac{t^2}2&\to\frac{x^2}2-\cdots\\ \frac{t^3}6&\to\frac{x^3}6-\cdots \end{align}$$
and in total
$$1+x+\frac{x^2}2+0\,x^3.$$
With a little more courage, up to degree $5$:
$$\begin{align}1&\to1\\ t&\to x-\frac{x^3}6+\frac{x^5}{120}\\ \frac{t^2}2&\to\frac{x^2}2-\frac{x^4}6\cdots\\ \frac{t^3}{6}&\to\frac{x^3}{6}-\frac{x^5}{12}\cdots\\ \frac{t^4}{24}&\to\frac{x^4}{24}\cdots\\ \frac{t^5}{120}&\to\frac{x^5}{120}\cdots \end{align}$$
gives
$$1+x+\frac{x^2}2-\frac{x^4}8-\frac{x^5}{15}.$$
Hint: if $f(x)= e^{\sin x}$, compute $f', f''$ and $f'''$. Then the first 4 terms are
$f(0), f'(0)x, \frac{f''(0)}{2}x^2$ and $\frac{f'''(0)}{6}x^3.$