Building on the hint $u=x^2$.
$\sin(u)=u-\frac{u^3}{3!}+\frac{u^5}{5!}$
$\frac{\sin(u)}{u}=1-\frac{u^2}{3!}+\frac{u^4}{5!}$
Now back substituting in $x^2$
$\frac{\sin(x^2)}{x^2}=1-\frac{x^4}{3!}+\frac{x^8}{5!}+...$
Edit
Using $x=x^2$ as the substitution as op had.
$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...$
$\frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}+...$
Now back substituting $x^2$
$\frac{\sin(x^2)}{x^2}=1-\frac{(x^2)^2}{3!}+\frac{(x^2)^4}{5!}...$
It doesn't matter what you pick as your substitution. It may get confusing if you use the substitution $x=x^2$.
This can be made a lot simpler by changing variables. (Changing variables is commonly taught as a technique for integration, but it can also be handy for differentiation.)
Introduce the new variable $u=1-x$. Then $x=1-u$, and
$$f(x) = \frac{1-u}{\sqrt{u}} = u^{-1/2} - u^{1/2}$$
If we define a new function $g(x)=x^{-1/2} - x^{1/2}$ then this tells us that $$f(x) = g(1-x),$$ and therefore on taking derivatives we have $$f^{(n)}(x) = (-1)^n g^{(n)}(1-x)$$
This change of variables allows you to essentially swap out the problem of computing derivatives of $f(x)$ and trade it for computing derivatives of the (much simpler) function $g(x)$.
Now, the derivatives of $g(x)$ are
$$g'(x) = \left( - \frac{1}{2}\right)x^{-3/2} - \left(\frac{1}{2}\right)x^{-1/2} $$
$$g''(x) = \left( - \frac{1}{2}\right)\left( - \frac{3}{2}\right)x^{-5/2} - \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)x^{-3/2} $$
$$g'''(x) = \left( - \frac{1}{2}\right)\left( - \frac{3}{2}\right)\left( - \frac{5}{2}\right)x^{-7/2} - \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)x^{-5/2} $$
and in general if we introduce the notation $A_n$ to denote the product of the first $n$ odd numbers (for example, $A_1=1$, $A_2 = 1\cdot 3$, $A_3 = 1\cdot 3 \cdot 5$, etc.) then
$$g^{(n)}(x)=(-1)^n \frac{A_n}{2^n}x^{-(2n+1)/2} + (-1)^n \frac{A_{n-1}}{2^n}x^{-(2n-1)/2}$$
Now we recall that $f^{(n)}(x) = (-1)^n g^{(n)}(1-x)$, so that
$$f^{(n)}(x)=\frac{A_n}{2^n}(1-x)^{-(2n+1)/2} + \frac{A_{n-1}}{2^n}(1-x)^{-(2n-1)/2}$$
The only thing left is to express the coefficients $A_n$ in a more convenient closed form; for that, see Proving formula for product of first n odd numbers.
Best Answer
using serie expansion $$let : \;f(x)=(1+2x^3)e^{x^2}$$ $$ e^{x^2}＝1＋x^2＋\frac {x^4}{2!}＋{\rm O}(x^5)\\~\\f(x)=(1＋2x^3)e^{x^2}＝(1＋2x^3)(1＋x^2＋\frac {x^4}{2!})＋＋{\rm O}(x^5)＝1＋x^2+2x^3+\frac {1}{2}x^4＋2x^5+{\rm O}(x^5)$$
And therfore : $$f(x)=f(0)＋\frac {f^{(1)}(0)}{1!}x＋\frac {f^{(2)}(0)}{2!}x^2＋\frac {f^{(3)}(0)}{3!}x^3＋\frac {f^{(4)}(0)}{4!}x^4＋\frac {f^{(5)}(0)}{5!}x^5＋{\rm O}(x^5)$$
We note that: $$\frac {f^{(5)}(0)}{5!}＝2\implies\;f^{(5)}(0)＝2×5!=240$$
So, the fifth-degree derivative at x=0 is:240