I want to find out the fourth degree Maclaurin expansion of the following function.
$$f(x)=\sin\left(x^2\right)\tag{1}$$
$$\frac{\mathrm{d}^{0}}{\mathrm{d^{0}x}}f(x)=\sin\left(x^2\right)~\implies~f^{(0)}(0)=0\tag{2}$$
$$\frac{\mathrm{d}^{1}}{\mathrm{d^{1}x}}f(x)=2x\cos\left(x^2\right)~\implies~f^{(1)}(0)=0\tag{3}$$
$$\frac{\mathrm{d}^{2}}{\mathrm{d^{2}x}}f(x)=2\left(\cos\left(x^2\right)+x\left(-\sin\left(x^2\right)2 x\right)\right)\tag{4}$$
$$=2\left(\cos\left(x^2\right)-2x^2\sin\left(x^2\right)\right)~\implies~f^{(2)}(0)=2\tag{5}$$
$$\frac{\mathrm{d}^{3}}{\mathrm{d^{3}x}}f(x)=2\left(-2x\sin\left(x^2\right)-2\left(2x\sin\left(x^2\right)+x^2\cdot 2 x\cos\left(x^2\right)\right)\right)\tag{6}$$
$$=2\left(-2x\sin\left(x^2\right)-4 x\sin\left(x^2\right)-4x^3\cos\left(x^2\right)\right)\tag{7}$$
$$=-4x\sin\left(x^2\right)-8x\sin\left(x^2\right)-8 x^3\cos\left(x^2\right)\tag{8}$$
$$=-12x\sin\left(x^2\right)-8x^3\cos\left(x^2\right)~\implies~f^{(3)}(0)=0\tag{9}$$
$$\frac{\mathrm{d}^{4}}{\mathrm{d^{4}x}}f(x)=-12\left(\sin\left(x^2\right)+x\cdot 2x\cos\left(x^2\right)\right)-8\left(3x^2\cos\left(x^2\right)+x^3\cdot 2x\left(-\sin\left(x^2\right)\right)\right)\tag{10}$$
$$=\underbrace{16x^4\sin\left(x^2\right)-48 x^2\cos\left(x^2\right)-12\sin\left(x^2\right)}_{\text{Ommited detailed derivation}}~\implies~f^{(4)}(0)=0\tag{11}$$
$$f(x)=\sum_{i=0}^{\infty}{f^{(i)}(0)\over i!}x^i\tag{12}$$
$$={f^{(0)}(0)\over 2!}x^2+\sum_{i=5}^{\infty}{f^{(i)}(0)\over i!}x^i\tag{13}$$
$$=x^2+\sum_{i=5}^{\infty}{f^{(i)}(0)\over i!}x^i\tag{14}$$
What I can't get is that as$~x~$is very small, then the following can be held.
$$\color{red}{f(x)\approx x^2}\tag{15}$$
I can guess$~5\leq\text{ith}~$derivative of$~f(x)~$with$~x=0~$is always$~0~$
But I don't have any clue to prove it, and moreover as it is held then symbol$~\approx~$is unnecesary.
Then why the approximation can be taken?
You showed (quite laboriously) that $\sin(x^2)$ is equal to $x^2$ plus some terms involving $x^5$ and higher powers. When $x$ is small, $x^5$ is much smaller than $x^2$ so you can just ignore the higher power terms and say that $\sin(x^2) \approx x^2$.