So using chain rule for the first derivative:
$$f' = 2x\cdot e^{x^2}$$
then using product rule for the $$f'' = 2x(2xe^{x^2}) + e^{x^2}\cdot 2$$
Is there an easy way to get the third? Does the second deriative simplify to:
$$4x^2e^{x^2} + 2e^{x^2} = (4x^2 + 2)(e^{x^2})$$
Can I then take chain rule again to get $f^{(3)}$?
idk if that would be a interesting way, but you could call (using the fact it $e^{x^2}$ will never vanish from and get) $$f^{(n)}(x)=a_n(x)e^{x^2}$$ where the exponent denote the n-th derivate, then you derive the relation with the next term $$f^{(n+1)}(x)=\frac{d}{dx}\left(a_n(x)e^{x^2}\right)\\ =a'_n(x)e^{x^2}+2a_n(x)xe^{x^2}\\ =(a'_n(x)+2a_n(x)x)e^{x^2}=a_{n+1}(x)e^{x^2}\\ a_{n+1}(x)=a'_n(x)+2a_n(x)x$$ starting from $$n=0\Rightarrow f^{(0)}(x)=f(x)=e^{x^2}\Rightarrow a_0(x)=1$$ we get $$n=1\Rightarrow a_1(x)=(1)'+2(1)x=2x\\ n=2\Rightarrow a_2(x)=(2x)'+2(2x)x=2+4x^2\\ n=3\Rightarrow a_3(x)=(2+4x^2)'+2(2+4x^2)x=12x+8x^3\\ n=4\Rightarrow a_4(x)=(12x+8x^3)'+2(12x+8x^3)x=12+48x^2+16x^4$$ wich lead to $$f''''(x)=(12+48x^2+16x^4)e^{x^2}$$
It's all a matter of applying the Chain Rule and the Product Rule correctly.
$$f'(x)=2xe^{x^2}\\f''(x)=2e^{x^2}+4x^2e^{x^2}\\f'''(x)=4xe^{x^2}+8xe^{x^2}+8x^3e^{x^2}=e^{x^2}\left(12x+8x^3\right)\\f^{(4)}(x)=e^{x^2}(24x^2+16x^4)+e^{x^2}(12+24x^2)=e^{x^2}(12+48x^2+16x^4)$$
It could be interesting to use logarithmic differentiation $$y=e^{x^2}\implies\log(y)=x^2\implies \frac{y'}y=2x\implies y'=2xe^{x^2}$$ $$\log(y')=\log(2)+\log(x)+{x^2}\implies \frac{y''}{y'}=\frac 1x+2x\implies y''=(2+4x^2)e^{x^2}$$ $$\log(y'')=\log(2+4x^2)+x^2\implies \frac{y'''}{y''}=\frac{8x}{2+4x^2}+2x\implies y'''=4 x \left(2 x^2+3\right)e^{x^2}$$ $$\log(y''')=\log(4)+\log(x)+\log\left(2 x^2+3\right)+x^2\implies\frac{y''''}{y'''}=\frac 1x+\frac{4x}{3+2x^2}+2x$$ and so on.
you can use the relation between the derivatives to find any derivative $$y'=2xe^{x^2}\rightarrow y'=2(xy+0)$$ $$y''=2y+2xy'\rightarrow y''=2(xy'+y)$$ $$y'''=2(xy''+2y')$$ $$y''''=2(xy'''+3y'')$$ $$y'''''=2(xy''''+4y''')$$
Use Maclaurin expansion: $y=e^{x^2}=\sum_{n=0}^{\infty} \frac{x^{2n}}{n!}$: $$\begin{align}y'&=\sum_{n=1}^{\infty} \frac{2nx^{2n-1}}{n!}\\ y'' &=\sum_{n=1}^{\infty} \frac{2n(2n-1)x^{2n-2}}{n!}\\ y''' &=\sum_{n=2}^{\infty} \frac{2n(2n-1)(2n-2)x^{2n-3}}{n!}\\ y^{(4)}&=\sum_{n=2}^{\infty} \frac{2n(2n-1)(2n-2)(2n-3)x^{2n-4}}{n!}=\\ &=4\sum_{n=2}^{\infty} \frac{(2n-1)(2n-3)x^{2(n-2)}}{(n-2)!}=\\ &=4\sum_{n=0}^{\infty} \frac{(2n+3)(2n+1)x^{2n}}{n!}=\\ &=16\sum_{n=1}^{\infty} \frac{(n-1+1)x^{2n}}{(n-1)!}+32\sum_{n=1}^{\infty} \frac{x^{2n}}{(n-1)!}+12\sum_{n=0}^{\infty} \frac{x^{2n}}{n!}=\\ &=(16x^4+48x^2+12)e^{x^2}.\end{align}$$
