This is a vague step that I can't understand particularly.
Suppose we have the equation
$$x^3y''-y=0,$$ with solution $y \sim x^{\frac{3}{4}}\exp({2x^{-\frac{1}{2}}})\ $.
Then we make the transformation $$ y(x) = x^{\frac{3}{4}}\exp({2x^{-\frac{1}{2}}})w(x)\ . $$
Then we are able to say that $w(x)$ satisfies $$w''+(\frac{3}{2x}-\frac{2}{x^\frac{3}{2}})w'-\frac{3w}{16x^2}=0$$
Now I am certain it's quite simple as it is seen as a minor predecessor step for more important things however, I'm not entirely sure how we see that $w(x)$ would satisfy that equation.. any help is greatly appreciated.
Why, you just calculate the derivatives:
\begin{align} \ & y=x^{\frac 34}\exp (2x^{-\frac 12})w(x) \\ \ \implies & y''=\biggl[x^{\frac 34}w''(x)+2\Bigl(\frac 34x^{-\frac 14}-x^{-\frac 34}\Bigl)w'(x) +\Bigl(x^{-\frac 94}-\frac{3}{16}x^{-\frac 54}\Bigl)w(x) \biggl]\exp (2x^{-\frac 12}) \\ \end{align}
And then plug it into the original equation:
\begin{align} \ & x^3y''-y=0 \\ \ \implies & \biggl[x^{\frac{15}{4}}w''(x)+2(\frac 34x^{\frac{11}{4}}-x^{\frac 94})w'(x) +(x^{\frac 34}-\frac{3}{16}x^{\frac 74})w(x) \biggr]-\biggl[x^{\frac 34}w(x)\biggr]=0 \\ \ \implies & x^{\frac{15}{4}}w''(x)+(\frac 32x^{\frac{11}{4}}- 2x^{\frac 94})w'(x) -\frac{3}{16}x^{\frac 74}w(x) =0 \\ \ \implies & w''(x)+(\frac 32x^{-1}- 2x^{-\frac 32})w'(x) -\frac{3}{16}x^{-2}w(x) =0 \\ \ \implies & w''(x)+(\frac{3}{2x}-\frac{2}{x^{\frac 32}})w'(x)-\frac{3}{16x^2}w(x)=0 \end{align}