Rewriting the time-independent Schrödinger equation for a simple harmonic oscillating potential in terms of new variables Show that the time-independent Schrödinger equation for a simple harmonic oscillating potential 
$$-\frac{\hbar^2}{2m}\frac{d^2 u}{dx^2}+\frac12 m\,\omega_0^2 x^2u=E\,u$$ can be written as 
$$\frac{d^2u}{dy^2}+(\alpha - y^2)u=0\tag{1}$$
where $$y=\sqrt{\frac{m\,\omega_0}{\hbar}}x$$
and $$\alpha=\frac{2E}{\hbar\,\omega_0}$$

So by my logic $$\frac{dy}{dx}=\sqrt{\frac{m\,\omega_0}{\hbar}}$$ and 
$$\frac{d^2y}{dx^2}=0$$
clearly something has gone wrong or I am going about this the wrong way. My lecturer mentioned in the lecture that: 

From $$y=\sqrt{\frac{m\,\omega_0}{\hbar}}x\tag{2}$$ it follows that $$\fbox{$\frac{d}{dx}=\sqrt{\frac{m\,\omega_0}{\hbar}}\frac{d}{dy}$}\tag{3}$$

But how does $(3)$ follow from $(2)$? 
I know that the boxed equation $(3)$ is correct as I have checked the printed notes for that lecture (for which the relevant parts are shown below):


But that's just the first problem; even if I understood how obtain $(3)$, I still don't understand how to proceed to derive $(1)$. 
The only thing I could think to do is take the second derivative of $(3)$ with respect to $x$ such that
$$\frac{d}{dx}\frac{d}{dx}=\frac{d^2}{dx^2}=\frac{d}{dx}\sqrt{\frac{m\,\omega_0}{\hbar}}\frac{d}{dy}$$ and once again I am left confused as this approach doesn't appear to be getting me any closer to deriving $(1)$.
I don't doubt that I am likely missing something very simple here, but at the moment I have no idea what that is. 
Could someone please provide me with some hints or advice on how to derive $(1)$ and what operations where carried out to obtain $(3)$?

Kindest regards.
 A: Your first mistake was
$$
\frac{dy}{dx} = c
$$
what you should be writing 
$$
\frac{d}{dx} = \frac{dy}{dx}\frac{d}{dy}
$$
quite a difference.
To apply this twice we compute
$$
\frac{d}{dx}\frac{du}{dx} = \frac{dy}{dx}\frac{d}{dy}\left(\frac{dy}{dx}\frac{du}{dy}\right) 
$$
which using $c = \frac{dy}{dx}$ we can compute
$$
\frac{d^2u}{dx^2} = c\frac{d}{dy}\left( c\frac{du}{dy}\right) = c^2\frac{d^2u}{dy^2}
$$
which should work out to what you require.
Now you may ask, why do I use $\frac{dy}{dx} = c$ in the latter expression when I stated that this was wrong ? Well, the change of variables (as per the comments) does contain the derivative $\frac{dy}{dx}$ which is the straight forward derivative of the expression $y = cx$.
A: First    of  all: Your logic is sound. Since 
\begin{align*}
y=\sqrt{\frac{m\,\omega_0}{\hbar}}x
\end{align*}
we obtain
\begin{align*}
\frac{dy}{dx}=\sqrt{\frac{m\,\omega_0}{\hbar}}
\end{align*}

We consider starting from $u=u(x)$ and $y=y(x)$ the function $u=u(y(x))$ and obtain following  variations of the same theme
  \begin{align*}
\left(u(y(x))\right)^\prime&=u^\prime(y(x))y^\prime(x)\\
\frac{d}{dx}u(y(x))&=\frac{d}{dy}u(y)\frac{d}{dx}y(x)\\
\frac{du}{dx}&=\frac{du}{dy}\cdot\frac{dy}{dx}\\
\frac{du}{dx}&=\frac{du}{dy}\sqrt{\frac{m\,\omega_0}{\hbar}}\tag{1}\\
\end{align*}
  or just looking at  the operator $\frac{d}{dx}$ without explicitly addressing the function $u$ the operator is to apply:
  \begin{align*}
\frac{d}{dx}&=\sqrt{\frac{m\,\omega_0}{\hbar}}\frac{d}{dy}
\end{align*}
  This explains the transition from (2) to (3).

$$ $$

The time-independent Schrödinger equation:
In order to normalise the equation we multiply according to the lecture notes  both sides with $\frac{2}{\omega_0 \hbar}$ and obtain with $\alpha=\frac{2E}{\hbar\,\omega_0}$
  \begin{align*}
\frac{2}{\omega_0 \hbar}\left(-\frac{\hbar^2}{2m}\frac{d^2 u}{dx^2}+\frac12 m\,\omega_0^2 x^2u\right)&=\left(\frac{2}{\omega_0 \hbar}\right)E\,u\\
-\frac{\hbar}{\omega_0m}\frac{d^2 u}{dx^2}+\frac{m\,\omega_0}{\hbar} x^2u&=\alpha u\\
-\frac{\hbar}{\omega_0m}\frac{d^2 u}{dx^2}+y^2u&=\alpha u\\
\frac{\hbar}{\omega_0m}\frac{d^2 u}{dx^2}+\left(\alpha  -y^2\right)u&=0\tag{2}\\
\end{align*}

Finally, since $u=u(y(x))$ we obtain from (1)
\begin{align*}
\frac{d^2 u}{dx^2}&=\frac{d}{dx}\left(\frac{du}{dx}\right)\\
&=\frac{d}{dx}\left(\frac{du}{dy}\cdot\frac{dy}{dx}\right)\\
&=\frac{d}{dx}\left(\frac{du}{dy}\cdot\sqrt{\frac{m\,\omega_0}{\hbar}}\right)\\
&=\sqrt{\frac{m\,\omega_0}{\hbar}}\cdot\frac{d}{dx}\left(\frac{du}{dy}\right)\tag{3}\\
&=\sqrt{\frac{m\,\omega_0}{\hbar}}\cdot\frac{d^2u}{dy^2}\cdot\frac{dy}{dx}\\
&=\sqrt{\frac{m\,\omega_0}{\hbar}}\cdot\frac{d^2u}{dy^2}\cdot\sqrt{\frac{m\,\omega_0}{\hbar}}\\
&=\frac{m\,\omega_0}{\hbar}\cdot\frac{d^2u}{dy^2}\\
\end{align*}

and we conclude from (2)
  \begin{align*}
\frac{d^2u}{dy^2}+(\alpha - y^2)u=0
\end{align*}
Observe in (3) we apply $\frac{d}{dx}$ to $\frac{du}{dy}$ similarly as we did before when applying $\frac{d}{dx}$ to $u$.

