If $P(x)$ is a polynomial of degree three in $x$, and $y^2 = P(x)$, show that $\frac{D(y^3D^2y)}{y^2}$ is constant If $P(x)$ is a polynomial of degree three in $x$, and $y^2 = P(x)$, show that
$$\frac{D(y^3D^2y)}{y^2}$$
is a constant, where D denotes the derivative operator.
I have tried expressing the expression above in terms of $P$ and its derivatives (in the hopes of showing the derivative is $0$) but I couldn't manage to do that.

First consider
$$Dy = \frac{1}{2y}Dy^2$$
and
$$D^2y = D(\frac{1}{2y}Dy^2) =\frac{-1}{2y^2}(Dy)(Dy^2) + \frac{1}{2y}D^2y^2 = -\frac{1}{4y^3}(Dy^2)^2 + \frac{1}{2y}D^2y^2 $$
Therefore, the numerator in your expression is
\begin{align}
D(y^3D^2y) &= D(-\frac{1}{4}(Dy^2)^2 + \frac{y^2}{2}D^2y^2)
\\
&=-\frac{1}{2}(Dy^2)(D^2y^2) + \frac{1}{2}(Dy^2)(D^2y^2) + \frac{y^2}{2}D^3y^2
\\
&=\frac{y^2}{2}D^3y^2
\end{align}
then your original expression becomes
$$\frac{D(y^3D^2y)}{y^2} = \frac{1}{2}D^3 y^2$$
If $y^2 = ax^3+bx^2+cx+d$, then
$$\frac{D(y^3D^2y)}{y^2} = 3a$$
