Derivate by $x^2$

Call your variable $\;t:=x^2\implies \sqrt t=x\;,\;\;t\ge0\;$ , so you want

$$\frac d{dx^2}(x)=\frac d{dt}(\sqrt t)=\frac1{2\sqrt t}=\frac1{2x}$$

Let $f(x^2)=x$.

Hence, for $x>0$ $$f'(x^2)=(\sqrt{x^2})'_{x^2}=\frac{1}{2\sqrt{x^2}}=\frac{1}{2x}$$

If $x<0$ so $$f'(x^2)=-(\sqrt{x^2})'_{x^2}=-\frac{1}{2\sqrt{x^2}}=\frac{1}{2x}$$

$$\large \frac{df(x)}{dg(x)}=\large \frac{\frac{df(x)}{dx}}{\frac{dg(x)}{dx}}=\frac{f'(x)}{g'(x)}$$ for example $$f(x)=x^4+x^2+1 ,g(x)=x^2 $$ $$\frac{df}{dg}=\frac{d(x^4+x^2+1)}{d(x^2)}=\frac{4x^3+2x}{2x}=2x^2+1$$ if you take $u=x^2$ you need to find $\frac{df}{du}$ so first rewrite $f(x)$ by $u$ $$f(x)=(x^2)^2+x^2+1=u^2+u+1 \\ \to \frac{df}{du}=2u+1=2x^2+1$$ another example :$$f(x)=sinx ,g(x)=tan x \\\frac{df}{dg}=\frac{cos x}{tan^2 x+1}=sec^3x$$

another example : $$\frac{dx}{d(lnx)}=?\\=\frac{1}{\frac{1}{x}}=x$$