Follow the instructions
a) $x=e^t,~z(t)=y(x(t))$
I will use upper dot notation to denote time derivatives and prime notation to denote derivatives of $y$ with respect to $x$ I.e.
$\dot{z} = \dfrac{dz}{dt}$
$y^\prime = \dfrac{dy}{dx}$
$\dot{z} = y^\prime \dot{x} = y^\prime e^t$
$y^\prime = \dot{z} e^{-t}$
$\begin{align*}
&\ddot{z}= y^{\prime\prime}(\dot{x}^2) + y^\prime \dot{x} \\
&=y^{\prime\prime} e^{2t} + y^\prime e^t \\
&=y^{\prime\prime} e^{2t} + \dot{z}
\end{align*}$
$y^{\prime\prime} = (\ddot{z}-\dot{z})e^{-2t}$
Now you plug $y^\prime, ~y^{\prime\prime}$ into the original differential equation, letting $x=e^t$, and solve it in $t$
Finally let $t = \ln(x)$
$\dot{z} = y^\prime \dot{x}$
$\begin{align*}
&\dfrac{d}{dt} \dot{z} = \\
&\left(\dfrac{d}{dt} y^\prime\right)\dot{x} + y^\prime \ddot{x} = \\
&\left(\dfrac{d}{dx} y^\prime \dot{x}\right)\dot{x} + y^\prime \ddot{x} = \\
&\left(y^{\prime\prime}\dot{x}\right)\dot{x} + y^\prime \ddot{x} = \\
&y^{\prime\prime}\dot{x}^2 + y^\prime \ddot{x}
\end{align*}$