Differentiate $y=x$ with respect to $x^2$? Let $y=x$. Differentiate $y$ with respect to $x^2$.


My intuition tells me to go down a substitution route here. So, let $u=x^2~~ \Leftrightarrow~~x=\sqrt{u}$, then what we are looking to find is simply

$$\frac{d}{du}\sqrt{u}$$
which is
$$\frac{1}{2\sqrt{u}}$$
as $\sqrt{u}=x$,
$$\frac{d}{dx^2}~x=\frac{1}{2x}$$

is this approach/solution correct, or is there something fundamentally wrong here? Any help is appreciated.
Yes it is correct. 

Alternatively, from the Chain Rule, $$y=x\implies y^2=x^2\implies \frac{dy^2}{d(x^2)}=\frac{d}{d(x^2)}(x^2)\implies 2y\frac{dy}{d(x^2)}=1\implies \boxed{\frac{dy}{dx^2}=\frac1{2y}=\frac1{2x}}$$

Note that this method is used very often in statistics (e.g. maximum likelihood estimates for the variance since we estimate $\sigma^2$ rather than $\sigma$; hence we prefer to differentiate the former).
Set $z=x^2$. Then $x= \sqrt z$ and:
$$\frac{d}{d(x^2)}y=\frac{d}{d(x^2)}x=    \frac{d}{dz}\sqrt z=   \frac{1}{2 \sqrt z}=  \frac{1}{2x}  $$