My professor used to say:
You might want to do calculus in $\Bbb{R}$, but the functions themselves naturally live in $\Bbb{C}$. Euler was the first to discover that if you don't look at what they do everywhere in the complex plane, you don't really understand their habits.
This is as subjective as it gets, but it has always helped my intuition. In particular, you might think that some function is doing nothing wrong, so it should be analytic. Well, if it does nothing wrong in $\Bbb{R}$, look at what it does in $\Bbb{C}$! If also in $\Bbb{C}$ it does nothing wrong, then it is analytic. If in $\Bbb{C}$ it makes some mess, then you have to be careful also in $\Bbb{R}$. To quote my professor again:
Even in $\Bbb{R}$, and in the most practical and applied problems, you can hear distant echos of the complex behavior of the functions. It's their nature, you can't change it.
Fix $u>0$. Maclaurin expansion $\sin(u)$ around $u$ is:
$$\sin(u)=\sum_{n=0}^{\infty}\frac{(-1)^nu^{2n+1}}{(2n+1)!}$$
Now $u$ is fixed, so you can express $u$ as $u=x^2$ for some $x$ and rewrite series:
$$\sin(x^2)\sin(u)=\sum_{n=0}^{\infty}\frac{(-1)^nu^{2n+1}}{(2n+1)!}=\sum_{n=0}^{\infty}\frac{(-1)^n(x^2)^{2n+1}}{(2n+1)!}$$
Note that it is Taylor series for $\sin(x^2)$
If you have Taylor series for some function $f(x)$ around $x_0\neq 0$, you also can substitute $u=x^2$, but you don't get Taylor series as result:
$$f(u)=\sum_{n=0}^{\infty}a_n(u-x_0)^n$$
If you substitute $u=x^2$ in this case you get:
$$f(x^2)=\sum_{n=0}^{\infty}a_n(x^2-x_0)^n$$
It's something, but it is not Taylor series in form:
$$f(x^2)=\sum_{n=0}^{\infty}b_n(x-x_0)^n$$
Best Answer
The general formula would be $\frac{(-1)^{0+1} (-1)!}{2^0} = -(-1)!$. The factorial of $-1$ is undefined, so the general formula isn't defined for $n=0$. However $f^{(0)}(2) = f(2) =\ln 2$. That's what is meant by "not working".
Note that this is closely related to $\int x^{n-1}\ \mathrm dx = \frac1n x^n$ wich is also not defined for $n=0$ but $\int x^{-1} \ \mathrm dx = \ln x$.