How To Use The Steps on Page 149 in Calculus Made Easy To Solve Chapter XIV Example 12 Please how do I take the derivative of
$ y = \left(\frac{1}{a^x}\right)^{ax} $
from Calculus Made Easy Chapter XIV Example 12 using the steps used 
to solve 
$y=a^x$ 
on page 149.
The steps are 
\begin{align*}
y & = a^x\\
\log_ey & = x\log_e a\\
x & =\frac{\log_ey}{\log_ea} = \frac{1}{\log_ea}\, x \,\log_ey\\
\frac{dx}{dy} & =\frac{1}{\log_ea}\, x \, \frac{1}{a^2 \, x\, \log_ea}\\
\frac{dy}{dx} & = \frac{1}{\frac{dx}{dy}} = a^x \, x\, \log_ea
\end{align*}
These were my steps and were I got stuck
\begin{align*}
y & = \left(\frac{1}{a^x} \right)^{ax}\\
y & = a^{-ax^2}\\
\log_ey & = \log_ea^{-ax^2}\\
\log_ey & = -ax^2\log_ea\\
\frac{\log_ey}{\log_ea} & = -ax^2
\end{align*}

A good place to start is to consider that
$$\bigg(\dfrac{1}{a^x} \bigg)^{ax} = (a^{-x}) ^{ax} = a^{-ax^2}.$$
Once you get there, apply logarithmic differentiation to get
\begin{array}[rcl] 
$y & =&a^{-ax^2}\\
\ln y& =& -ax^2 \ln a\\
\dfrac{1}{y}\cdot\dfrac{dy}{dx}&=&-2a\ln a \cdot x\\
\dfrac{dy}{dx}&=& -2a\ln a \cdot a^{-ax^2}\cdot x
\end{array} 
...assuming you are solving for the derivative of y with respect to x. 
Edit:  Following the steps in your book, we can do the following:
\begin{array}[rcl]
$y & = & a^{-ax^2} \\
\ln y & = & -ax^2 \ln a \\
-ax^2 & = & \dfrac{1}{\ln a}\cdot \ln y \\
-2ax \cdot \dfrac{dx}{dy} & = & \dfrac{1}{\ln a}\cdot \dfrac{1}{y} \\
\dfrac{dx}{dy} & = & -\dfrac{1}{2ax\cdot\ln a}\cdot \dfrac{1}{y} \\
\dfrac{dy}{dx} & = & -2ax\cdot \ln a\cdot y \\
\dfrac{dy}{dx} & = &-2a\ln a\cdot a^{-ax^2}\cdot x \\
\end{array}
