$$
\cos ^{-1}\left(\frac{y}{b}\right)=2 \log \left(\frac{x}{2}\right), x>0
$$
On differentiating w.r.t. $x$ we get
$$
\begin{array}{cc}
& -\frac{1}{\sqrt{1-\frac{y^2}{b^2}}} \cdot \frac{1}{b} \frac{d y}{d x}=2 \cdot \frac{1}{\left(\frac{x}{2}\right)} \cdot \frac{1}{2} \\
\Rightarrow \quad & -\frac{1}{\sqrt{b^2-y^2}} \cdot \frac{d y}{d x}=\frac{2}{x} \\
\Rightarrow \quad & x \frac{d y}{d x}=-2 \sqrt{b^2-y^2}
\end{array}
$$
Again, differentiating w.r.t. $x$ we get,
$$
\begin{aligned}
& x \frac{d^2 y}{d x^2}+\frac{d y}{d x}=-2 \cdot \frac{1}{2}\left(b^2-y^2\right)^{-1 / 2}(-2 y) \frac{d y}{d x} \\
\Rightarrow \quad & \quad x \frac{d^2 y}{d x^2}+\frac{d y}{d x}=\frac{2 y}{\sqrt{b^2-y^2}} \cdot \frac{d y}{d x} \\
\Rightarrow \quad & x \frac{d^2 y}{d x^2}+\frac{d y}{d x}=\frac{2 y}{-\frac{x}{2} \cdot \frac{d y}{d x}} \cdot \frac{d y}{d x} \quad \text { [from Eq. (i)] } \\
\Rightarrow \quad & x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}=-4 y
\end{aligned}
$$