Differentiation the form $\frac{dy}{dt}$ with respect to y.

Question

It's been awhile since I took differential equations, so I am unsure if my manipulation is correct. Isn't it true that if we set $h(t,y)=y,$ then $h_{ty}=h_{yt}$? This would imply $$\dfrac{\partial}{\partial y}\dfrac{\partial y}{\partial t}=\dfrac{\partial}{\partial t}\dfrac{\partial y}{\partial y}=\dfrac{\partial}{\partial t}0=0.$$

Answer 1

Without following your calculus, a straightforward method is : $$\frac{d}{dy}\left(\frac{dy}{dt}\right)=\frac{d}{dy}\left(\frac{1}{\frac{dt}{dy}}\right)=-\frac{\frac{d^2t}{dy^2}}{\left(\frac{dt}{dy}\right)^2}$$ because to differentiate wrt $y$ we have to define which function of $y$ (not of $t$) must be considered. Thus it is not $\frac{dy}{dt}$ because this is a function of $t$. It is $\left(\frac{dt}{dy}\right)^{-1}$ which is a function of $y$ as required.

Answer 2

Setting h(t,y) to y appears bogus.
When y is a function of t, one usually writes y = y(t).
Then $y_y = 1$, not the zero you claim, though $y_{yt}$ = 0.
$y_t = y'$ since y is a function of t.
$y_{ty} = y'_y$ doesn't happen because y' is not a function of y.